Eigenvalues, Eigenvectors, and Eigenspaces_ Part 1_ Attempt review _ eClass
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Question 1
Correct
Mark 10.00 out of 10.00
Eigenvalues and Eigenvectors
The definitions of eigenvalues and eigenvectors, along with basic examples, are presented in the video Eigenvalues and eigenvectors 1
.
Definition:
Let be an matrix. A scalar (
pronounced lambda
) is called an eigenvalue
of
if there is a non-zero vector such
that
Such a vector is called an eigenvector
of corresponding to the eigenvalue .
What this definition means is that if is an eigenvector, then multiplying with transforms in the simplest possible way: it just multiplies
it by a scalar.
Warning: The zero vector is
not
considered to be an eigenvector!
Example: Let .
-2
-2
0
-2
Correct answer, well done.
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Correct answer, well done.
Marks for this submission: 1.00/1.00.
Thus,
-2
is an of and 1
1
0
is an of corresponding to -2
. Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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-3
-4
-2
for any scalar .
eigenvalue
eigenvector
≠
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Correct answer, well done.
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Correct answer, well done.
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Thus, of .
Correct answer, well done.
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is not an eigenvector
A correct answer is .
A correct answer is , which can be typed in as follows: -2
A correct answer is , which can be typed in as follows: -2
A correct answer is: "eigenvalue"
A correct answer is .
A correct answer is: "eigenvector"
A correct answer is , which can be typed in as follows: -2
A correct answer is .
A correct answer is: "≠"
A correct answer is: "is not an eigenvector"
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Question 2
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Geometric Interpretation
Let be an matrix.
Geometrically, a non-zero vector in an eigenvector of if and only if is parallel to .
Based upon these pictures:
Is an eigenvector of ?
Is an eigenvector of ?
Is an eigenvector of ?
Multiplication by the matrix is a linear transformation. More precisely, the map defined by is a linear
transformation. So an eigenvector of is a non-zero vector which is mapped by to a scalar multiple of itself.
In Block 4, we noted that matrices and linear transformations between Euclidean spaces (that is, the spaces and ) are in fact equivalent
concepts: every linear transformation is a matrix transformation and vice versa. So we also have equivalent concepts of eigenvalues and
eigenvectors for linear transformations. More precisely, a scalar is called an eigenvalue of a linear transformation if there is
a non-zero vector such that , in which case the vector is called an eigenvector of corresponding to the eigenvalue
. Indeed, is an eigenvalue of if and only if it is an eigenvalue of the standard matrix of . The same statement applies to
eigenvectors as well.
Example:
Consider the linear operator
given by reflection about some line in .
Any normal vector of is an eigenvector of ( and of ) corresponding to the eigenvalue -1
and any direction vector
of is an eigenvector of ( and of ) corresponding to the eigenvalue 1
.
Yes
No
Yes
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Correct
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Eigenvalues, Eigenvectors, and Eigenspaces: Part 1: Attempt review | eClass
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Question 3
Correct
Mark 14.00 out of 14.00
Exercises:
(a) Let , , and . Show that and are eigenvectors of and find their corresponding eigenvalues.
2
2
2
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Correct answer, well done.
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Thus 2
.
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Therefore,
is an eigenvector of corresponding to the eigenvalue 2
.
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-6
-3
-3
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Correct answer, well done.
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Thus -3
.
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Therefore,
is an eigenvector of corresponding to the eigenvalue -3
.
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(b) Let , , and . Show that and are eigenvectors of and find their corresponding
eigenvalues.
0
0
0
0
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Thus 0
.
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Therefore,
is an eigenvector of corresponding to the eigenvalue 0
.
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-3
0
3
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Thus .
Therefore,
is an eigenvector of corresponding to the eigenvalue 1
.
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A correct answer is .
A correct answer is , which can be typed in as follows: 2
A correct answer is , which can be typed in as follows: 2
A correct answer is , which can be typed in as follows: 2
A correct answer is .
A correct answer is , which can be typed in as follows: -3
A correct answer is , which can be typed in as follows: -3
A correct answer is , which can be typed in as follows: -3
A correct answer is .
A correct answer is , which can be typed in as follows: 0
A correct answer is , which can be typed in as follows: 0
A correct answer is , which can be typed in as follows: 0
A correct answer is .
A correct answer is , which can be typed in as follows: 1
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Eigenvalues, Eigenvectors, and Eigenspaces: Part 1: Attempt review | eClass
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Question 4
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Mark 9.00 out of 9.00
How to Find the Eigenvalues of a Square Matrix
Question:
How do we compute the the eigenvalues of an \(n \times n\) matrix \(A\), that is, how do we find the values of the scalars \(\lambda\) for
which \(A\mathbf{x} = \lambda \mathbf{x}\), for some non-zero vector \(\mathbf{x} \in \mathbb{R}^n\)?
Let \(I\) be the \(n \times n\) identity matrix. The argument below is explained in the video Eigenvalues and eigenvectors 3
.
Consider the following:
\(\qquad \ A\mathbf{x} = \lambda \mathbf{x}\), for some \(\mathbf{x}\)
\( \mathbf{0} \)
\(\qquad \iff A\mathbf{x} - \lambda \mathbf{x} = \mathbf{0}\), for some \(\mathbf{x}\)
\(
\mathbf{0} \)
\(\qquad \iff (A - \lambda I) \mathbf{x} = \mathbf{0}\) has
\(\qquad \iff A - \lambda I \)
(By the Invertible Matrix
Theorem)
\(\qquad \iff \det(A - \lambda I) \)
(By the Invertible Matrix
Theorem)
Computing the determinant of \(A - \lambda I\) using cofactor expansion along any row or column produces a polynomial in the variable \
(\lambda\). (This will become more clear after working out some examples: if you want, you may look at the first example worked out in
Question 6 below and come back here afterwards.) Therefore, the eigenvalues of an \(n \times n\) matrix \(A\) are the solutions \(\lambda\) of
the polynomial equation \[\det(A - \lambda I) = 0\]
Definition:
For an \(n \times n\) matrix \(A\), \[C_A(\lambda) = \det(A - \lambda I)\] is a polynomial of degree \(n\) in the variable \(\lambda\)
called the characteristic polynomial
of \(A\).
So, once again we ask:
Question:
How do we compute the the eigenvalues of an \(n \times n\) matrix \(A\), that is, how do we find the values of the scalars \(\lambda\) for
which \(A\mathbf{x} = \lambda \mathbf{x}\), for some non-zero vector \(\mathbf{x} \in \mathbb{R}^n\)?
Answer:
To compute the eigenvalues of an \(n \times n\) matrix \(A\), we find the roots of the characteristic polynomial \(C_A(\lambda) = \det(A -
\lambda I)\) of \(A\).
We will go through some examples of how to compute \(\det(A - \lambda I)\) shortly, but first we will do some examples that involve finding
the eigenvalues of \(A\) once we know the characteristic polynomial \(C_A(\lambda)\) of \(A\).
Example: Suppose that \(A\) is a \(5 \times 5\) matrix with characteristic polynomial \[C_A(\lambda) = -(\lambda - 8)^3(\lambda + 6 )^2\]
Then \(A\) has two eigenvalues: \(\lambda = 8\) and \(\lambda =\) -6
.
Example:
Suppose that \(A\) is a \(7 \times 7\) matrix with characteristic polynomial \[C_A(\lambda) = -\lambda(\lambda + 5)(\lambda
≠
≠
a nontrivial solution.
is not invertible.
= 0
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+1)^2(\lambda - 2)^3\]
Then the eigenvalues of \(A\) are
.
Example: Suppose that \(A\) is a \(2 \times 2\) matrix with characteristic polynomial \[C_A(\lambda) = \lambda^2 + \lambda - 12\]
Then the eigenvalues of \(A\) are
.
Example:
Suppose that \(A\) is an \(n \times n\) matrix with characteristic polynomial \[C_A(\lambda) = (\lambda + 2)^2(\lambda - 7
)^5(\lambda - 9)\]
Then \(n =\) 8
(Hint: what is the degree of \(C_A(\lambda)\)?)
Note:
For a square matrix \(A\) with real entries, the characteristic polynomial \(C_A(\lambda)\) may have some roots in the field of complex
numbers, which would mean that \(A\) has complex eigenvalues. We will discuss complex numbers, complex eigenvalues and eigenvectors in
Block 6.
-5, -1, 0, 2
3, -4
Correct
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Question 5
Correct
Mark 7.00 out of 7.00
Algebraic Multiplicity
Definition:
The
algebraic multiplicity
of an eigenvalue \(\lambda_0\) of an \(n \times n\) matrix \(A\), denoted \(\operatorname{alg}
(\lambda_0)\), is the number of times \(\lambda_0\) appears as a root of the characteristic polynomial \(C_A(\lambda)\) of \(A\). Equivalently,
it is the number of times \(\lambda - \lambda_0\) appears as a factor of \(C_A(\lambda)\).
Example:
Suppose that \(A\) is a \(9 \times 9\) matrix which has characteristic polynomial \[C_A(\lambda) = -\lambda^3(\lambda -
2)^4(\lambda - 6)^2.\] Thus, the eigenvalues of \(A\) are \(0\), \(2\), and \(6\). These eigenvalues have algebraic multiplicities as follows:
\(\lambda = 0\) has algebraic multiplicity \(\operatorname{alg}(0) = 3\).
\(\lambda = 2\) has algebraic multiplicity \(\operatorname{alg}(2) = \) 4
.
\(\lambda = 6\) has algebraic multiplicity \(\operatorname{alg}(6) =\) 2
.
Example:
Suppose that \(A\) is a \(15 \times 15\) matrix which has characteristic polynomial \[C_A(\lambda) = -(\lambda + 5)^4(\lambda +
1)^5 (\lambda - 1)^2(\lambda - 11)(\lambda - 19)^3\]
The eigenvalue \(\lambda = \) 11
has algebraic multiplicity \(1\).
The eigenvalue \(\lambda = \) 1
has algebraic multiplicity \(2\).
The eigenvalue \(\lambda = \) 19
has algebraic multiplicity \(3\).
The eigenvalue \(\lambda = \) -5
has algebraic multiplicity \(4\).
The eigenvalue \(\lambda = \) -1
has algebraic multiplicity \(5\).
Correct
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Question 6
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Another example about finding the eigenvalues of a 3x3 matrix, which is a bit more complicated than the one found below, is presented in the
video Eigenvalues and eigenvectors 4
.
Example:
Let \(A = {\left[\begin{array}{ccc} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 2 & -2 & 0 \end{array}\right]}\). Compute the characteristic
polynomial \(C_A(\lambda)\) and eigenvalues of \(A\).
\( A - \lambda I = {\left[\begin{array}{ccc} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 2 & -2 & 0 \end{array}\right]} -
\begin{bmatrix}\lambda& 0 & 0 \\ 0 &
\lambda & 0\\ 0&0&\lambda \end{bmatrix} = \begin{bmatrix}-1 - \lambda & -1&1\\ 0 & -2 - \lambda & 0 \\ 2&-2&-\lambda \end{bmatrix}\)
Thus, \[\begin{aligned}C_A(\lambda) = \det(A - \lambda I) &= \begin{vmatrix}-1 - \lambda & -1&1\\ 0 & -2 - \lambda & 0 \\ 2&-2&-\lambda
\end{vmatrix}\\ \\ &= (-2 -\lambda) \begin{vmatrix}-1 - \lambda &1 \\ 2 & -\lambda\end{vmatrix} \hspace{2.2cm} \left(\begin{array}
{c} \text{expand}\\ \text{along} \; R_2 \end{array} \right)\\ \\&= -(\lambda + 2) \left((-1 -\lambda)(-\lambda) - 2\right) \\ \\ &= -(\lambda + 2)
(\lambda^2 + \lambda - 2)\\ \\& =
-(\lambda + 2)(\lambda + 2)(\lambda - 1) \\ \\ &= -(\lambda + 2)^2(\lambda - 1) \end{aligned}\]
Therefore, the characteristic polynomial of \(A\) is \(C_A(\lambda) = -(\lambda + 2)^2(\lambda - 1)\) ( or \(-\lambda^3 - 3\lambda^2 + 4\))
and the eigenvalues of \(A\) are \(\lambda = -2\) and \(\lambda = 1\).
The algebraic multiplicity of
\(\lambda = -2\) is \(\operatorname{alg}(-2) = \) 2
and the algebraic multiplicity of
\(\lambda = 1\) is \
(\operatorname{alg}(1) = \) 1
.
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Exercises:
(a) Let \(A = {\left[\begin{array}{cc} 4 & 1 \\ 11 & -6 \end{array}\right]}\). Find the characteristic polynomial and eigenvalues of \(A\).
\(A - \lambda I = {\left[\begin{array}{cc} 4 & 1 \\ 11 & -6 \end{array}\right]} - \begin{bmatrix}\lambda & 0\\0 & \lambda\end{bmatrix}
= \begin{bmatrix}4 - \lambda & 1\\11& -6- \lambda\end{bmatrix}\)
\(C_A(\lambda)= \det(A - \lambda I) = \begin{vmatrix}4 - \lambda & 1\\11& -6- \lambda\end{vmatrix} =
\lambda^2 + \) 2
\( \lambda \; - \)
35
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Correct answer, well done.
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1.00/1.00.
The eigenvalues of \(A\) are
-7
and 5
.
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(b) Let \(B = {\left[\begin{array}{ccc} 3 & 1 & -1 \\ -1 & 5 & -1 \\ -4 & 4 & 0 \end{array}\right]}\). Find the characteristic polynomial and
eigenvalues of \(B\).
\(C_B(\lambda)= \det(B - \lambda I) = -\lambda^3 + \) 8
\( \lambda^2 \; - \)
16
\(\lambda\)
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Correct answer, well done.
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The eigenvalues of \(B\) are
0
and 4
.
Correct answer, well done.
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A correct answer is \( 2 \), which can be typed in as follows: 2
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 2 \), which can be typed in as follows: 2
A correct answer is \( 35 \), which can be typed in as follows: 35
A correct answer is \( -7 \), which can be typed in as follows: -7
A correct answer is \( 5 \), which can be typed in as follows: 5
A correct answer is \( 8 \), which can be typed in as follows: 8
A correct answer is \( 16 \), which can be typed in as follows: 16
A correct answer is \( 0 \), which can be typed in as follows: 0
A correct answer is \( 4 \), which can be typed in as follows: 4
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Question 7
Correct
Mark 6.00 out of 6.00
Eigenvalues of Triangular Matrices
Example:
Find the characteristic polynomial and eigenvalues of \[A = {\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 & 4 & 0 & 0 \\ -1 & 5 & 2 &
0 \\ 0 & 3 & -6 & -7 \end{array}\right]}\]
\(A - \lambda I = {\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 & 4 & 0 & 0 \\ -1 & 5 & 2 & 0 \\ 0 & 3 & -6 & -7 \end{array}\right]}
-
\begin{bmatrix}\lambda& 0 & 0 &0 \\ 0 & \lambda & 0&0\\ 0&0&\lambda&0\\0&0&0&\lambda \end{bmatrix} = \begin{bmatrix}-
\lambda &0&0&0\\1&4 -\lambda&0&0\\-1&5&2-\lambda&0\\0&3&-6&-7 -\lambda\end{bmatrix}\)
Since \(A - \lambda I\) is a triangular matrix, its determinant is the of the entries on its main diagonal. Thus,
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\[C_A(\lambda) = \det(A - \lambda I) = -\lambda(4 - \lambda)(2 - \lambda)(-7 - \lambda)= \lambda(\lambda - 4)
(\lambda - 2)(\lambda + 7)\]
and the eigenvalues of \(A\) are
4
, 2
,
0
, and -7
.
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Did you notice anything about the eigenvalues of \(A\) in this case?
The eigenvalues of \(A\) are the entries
of \(A\).
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Theorem:
The eigenvalues of a triangular \(n \times n\) matrix \(A\) (upper, lower, or diagonal) are the entries on the main diagonal of \(A\).
Example: Let \(B = {\left[\begin{array}{cccc} 18 & 3 & -1 & 7 \\ 0 & -9 & 3 & 0 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 27 \end{array}\right]}\).
The eigenvalues of \(B\) are
18
, -9
,
5
, and 27
.
Correct answer, well done.
Marks for this submission: 2.00/2.00.
product
on the main diagonal
A correct answer is: "product"
A correct answer is \( 0 \), which can be typed in as follows: 0
A correct answer is \( 4 \), which can be typed in as follows: 4
A correct answer is \( 2 \), which can be typed in as follows: 2
A correct answer is \( -7 \), which can be typed in as follows: -7
A correct answer is: "on the main diagonal"
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Question 8
Correct
Mark 1.00 out of 1.00
A correct answer is \( 18 \), which can be typed in as follows: 18
A correct answer is \( -9 \), which can be typed in as follows: -9
A correct answer is \( 5 \), which can be typed in as follows: 5
A correct answer is \( 27 \), which can be typed in as follows: 27
How to Find Eigenvectors of a Square Matrix
Recall:
Let \(A\) be an \(n \times n\) matrix and let \(\lambda\) be an eigenvalue of \(A\). A non-zero vector \(\mathbf{x} \in \mathbb{R}^n\) is
an eigenvector of \(A\) corresponding to \(\lambda\) if \(A\mathbf{x} = \lambda \mathbf{x}\).
Definition:
Let \(A\) be an \(n \times n\) matrix and let \(\lambda\) be an eigenvalue of \(A\). The set of all eigenvectors of \(A\) corresponding
to \(\lambda\), together with the zero vector, is called the eigenspace
of \(A\) corresponding to \(\lambda\) and is denoted \(E_{\lambda}\).
More precisely,the eigenspace corresponding to \(\lambda\) is given by
\[E_{\lambda} = \operatorname{Nul}(A - \lambda I)\]
Indeed, \[\begin{aligned} \mathbf{x} \in E_{\lambda} & \iff A\mathbf{x} = \lambda \mathbf{x} \\ &\iff A\mathbf{x} - \lambda \mathbf{x} =
\mathbf{0} \\&\iff (A-\lambda I)\mathbf{x} = \mathbf{0} \\&\iff \mathbf{x} \in \operatorname{Nul}(A - \lambda I)\end{aligned}\]
In particular, since \(E_{\lambda}\) is the
of the matrix \(A - \lambda I\), it is a subspace of \(\mathbb{R}^n\).
Definition:
Let \(A\) be an \(n \times n\) matrix and let \(\lambda\) be an eigenvalue of \(A\). The geometric multiplicity
of \(\lambda\),
denoted \(\operatorname{geo}(\lambda)\), is the dimension of the eigenspace \(E_{\lambda}\) corresponding to \(\lambda\).
null space
Correct
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Question 9
Correct
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To describe explicitly any subspace, it is enough to find a basis. This is true, in particular, for eigenspaces. It is explained in the video
Eigenvalues and eigenvectors 2 how to find the basis of one eigenspace for a specific matrix. Another example is given below.
Example:
Let \(A = {\left[\begin{array}{ccc} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 2 & -2 & 0 \end{array}\right]}\). In Question 6, we determined that the
characteristic polynomial of \(A\) is \[C_A(\lambda) = -(\lambda + 2)^2(\lambda - 1)\] and that the eigenvalues of \(A\) are \(\lambda = 1\)
and \(\lambda = -2\). The matrix \(A\) has two eigenspaces \(E_{1}\) and \(E_{-2}\). We will compute a basis for each eigenspace.
For \(\lambda =1\), we have \(A - \lambda I = A - I \) and so \(E_{1} = \operatorname{Nul}(A - I) \).
\[ [A - I\, | \, \mathbf{0}\, ] = \left[\begin{array}{ccc|c}-2&-1&1&0\\0&-3&0&0\\2&-2&-1&0\end{array}\right] \rightarrow \cdots \text{row
operations} \cdots \rightarrow \left[\begin{array}{ccc|c}1&0&-\frac{1}{2}&0\\0&1&0&0\\0&0&0&0\end{array}\right] \]
The resulting system becomes \[\begin{aligned}x_1 &= \frac{1}{2}t \\x_2 &= 0\\x_3 &= t \; \; \; (\text{free})\;\; \; \; \; \; \; \; \;\; t\in \mathbb{R}
\end{aligned}\]So the general solution of \((A - I)\mathbf{x} = \mathbf{0}\) is \[ \mathbf{x} = t\begin{bmatrix}\frac{1}{2}\\0\\1\end{bmatrix}, \; \;
\; t\in \mathbb{R}\]Thus a basis for the eigenspace \(E_1\) is \[\mathcal{B}_{E_1} = \left \{ \begin{bmatrix}\frac{1}{2}\\0\\1\end{bmatrix}
\right\}\].
Algebraic multiplicity of \(\lambda = 1\): \(\operatorname{alg}(1) =\)
1
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Geometric multiplicity of \(\lambda = 1\): \(\operatorname{geo}(1) =\)
1
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Marks for this submission: 1.00/1.00.
Note:
To avoid fractions in our basis, we can replace the vector \(\begin{bmatrix}\frac{1}{2}\\0\\1\end{bmatrix}\) by the scalar multiple
\
(2\begin{bmatrix}\frac{1}{2}\\0\\1\end{bmatrix} = \begin{bmatrix}1\\0\\2\end{bmatrix}\), so that a basis for the eigenspace \(E_1\) is \
[\mathcal{B}_{E_1} = \left \{ \begin{bmatrix}1\\0\\2\end{bmatrix} \right\}\] (Why can we do this?)
For \(\lambda = -2\), we have \(A - \lambda I = A - (-2) I = A + 2I\) and so \(E_{-2} = \operatorname{Nul}(A + 2I) \).
\[ [A + 2I\, | \, \mathbf{0}\, ] = \left[\begin{array}{ccc|c}1&-1&1&0\\0&0&0&0\\2&-2&2&0\end{array}\right] \overrightarrow{\begin{array}
{c}R_3 - 2R_1 \end{array}} \left[\begin{array}{ccc|c}1&-1&1&0\\0&0&0&0\\0&0&0&0\end{array}\right] \]
The general solution of \((A + 2I)\mathbf{x} = \mathbf{0}\) is
\(\mathbf{x} = s\)
1
1
0
\(+ t\)
-1
0
1
\(s,t \in\mathbb{R}\)
Correct answer, well done.
Marks for this submission: 2.00/2.00.
Thus a basis for the eigenspace \(E_{-2}\) is
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3/27/24, 1:23 AM
Eigenvalues, Eigenvectors, and Eigenspaces: Part 1: Attempt review | eClass
https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15219468&cmid=7601910
16/18
\(\mathcal{B}_{E_{-2}} = \left\{ \begin{array}
{c}\\ \\ \\ \\ \end{array}\right. \)
1
1
0
,
-1
0
1
\(\left. \begin{array}{c}\\ \\ \\ \\
\end{array}\right\}\)
Correct answer, well
done.
Marks for this
submission: 2.00/2.00.
Algebraic multiplicity of \(\lambda = -2\): \(\operatorname{alg}(-2) =\)
2
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Geometric multiplicity of \(\lambda = -2\): \(\operatorname{geo}(-2) =\)
2
Correct answer, well done.
Marks for this submission: 1.00/1.00.
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( \left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] \).
A correct answer is \( 2 \), which can be typed in as follows: 2
A correct answer is \( 2 \), which can be typed in as follows: 2
3/27/24, 1:23 AM
Eigenvalues, Eigenvectors, and Eigenspaces: Part 1: Attempt review | eClass
https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15219468&cmid=7601910
17/18
Question 10
Correct
Mark 12.00 out of 12.00
Exercise:
Let \(A = {\left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array}\right]}\).
Verify that the characteristic polynomial of \(A\) is
\[C_A(\lambda) = -\lambda(\lambda - 1)(\lambda - 2)\]
Thus, the eigenvalues of \(A\) are \(\lambda = 0\), \(\lambda = 1\) and \(\lambda = 2\).
Find bases for the three eigenspaces \(E_0\), \(E_1\), and \(E_2\) and state the algebraic multiplicity and geometric multiplicity of each
eigenvalue of \(A\).
For \(\lambda =0\), we have \(A - \lambda I = A \) and so \(E_{0} = \operatorname{Nul}(A) \).
A basis for the eigenspace \(E_{0}\) is
\(\mathcal{B}_{E_{0}} = \left\{ \begin{array}{c}\\ \\ \\
\\ \end{array}\right. \)
0
0
1
\(\left. \begin{array}{c}\\ \\ \\ \\
\end{array}\right\}\)
Correct answer, well
done.
Marks for this submission:
2.00/2.00.
Algebraic multiplicity of \(\lambda = 0\): \(\operatorname{alg}(0) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Geometric multiplicity of \(\lambda = 0\): \(\operatorname{geo}(0) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
For \(\lambda =1\), we have \(A - \lambda I = A - I \) and so \(E_{1} = \operatorname{Nul}(A - I) \).
A basis for the eigenspace \(E_{1}\) is
\(\mathcal{B}_{E_{1}} = \left\{ \begin{array}{c}\\ \\ \\
\\ \end{array}\right. \)
-1
0
1
\(\left. \begin{array}{c}\\ \\ \\ \\
\end{array}\right\}\)
Correct answer, well
done.
Marks for this submission:
2.00/2.00.
Algebraic multiplicity of \(\lambda = 1\): \(\operatorname{alg}(1) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Geometric multiplicity of \(\lambda = 1\): \(\operatorname{geo}(1) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
For \(\lambda =2\), we have \(A - \lambda I = A - 2I \) and so \(E_{2} = \operatorname{Nul}(A - 2I) \).
A basis for the eigenspace \(E_{2}\) is
3/27/24, 1:23 AM
Eigenvalues, Eigenvectors, and Eigenspaces: Part 1: Attempt review | eClass
https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15219468&cmid=7601910
18/18
\(\mathcal{B}_{E_{2}} = \left\{ \begin{array}{c}\\ \\ \\
\\ \end{array}\right. \)
-2
-1
1
\(\left. \begin{array}{c}\\ \\ \\ \\
\end{array}\right\}\)
Correct answer, well
done.
Marks for this submission:
2.00/2.00.
Algebraic multiplicity of \(\lambda = 2\): \(\operatorname{alg}(2) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Geometric multiplicity of \(\lambda = 2\): \(\operatorname{geo}(2) =\)
1
Correct answer, well done.
Marks for this submission: 1.00/1.00.
A correct answer is \( \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \).
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] \).
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( \left[\begin{array}{c} -2 \\ -1 \\ 1 \end{array}\right] \).
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 1 \), which can be typed in as follows: 1
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