Let n be an integer. Let T € L(R³) be given by the matrix n-2 n-1 n-1 n A = n n n+1 n+1 n + 2, Observe that A is symmetric (hence T is self-adjoint). In the following steps, we shall find an orthonormal basis of eigenvectors for T. (Note that for convenience of notation, vectors are written as row vectors.) a.) Show that vo = (1, −2, 1) is an eigenvector of T with eigenvlaue λ = 0. b.) Let W = E(0, T). Let w₁ = (1,0,−1) and w₂ {w1₁, w₂} is an orthogonal basis of W. 0 - (-237²). 3n B = = c.) Verify that Tw₁ = −2w₂ and Tw2 = (3n)w2 - 3w₁. Let B = [T|w]ß. Conclude that (1, 1, 1). Show that B =
Let n be an integer. Let T € L(R³) be given by the matrix n-2 n-1 n-1 n A = n n n+1 n+1 n + 2, Observe that A is symmetric (hence T is self-adjoint). In the following steps, we shall find an orthonormal basis of eigenvectors for T. (Note that for convenience of notation, vectors are written as row vectors.) a.) Show that vo = (1, −2, 1) is an eigenvector of T with eigenvlaue λ = 0. b.) Let W = E(0, T). Let w₁ = (1,0,−1) and w₂ {w1₁, w₂} is an orthogonal basis of W. 0 - (-237²). 3n B = = c.) Verify that Tw₁ = −2w₂ and Tw2 = (3n)w2 - 3w₁. Let B = [T|w]ß. Conclude that (1, 1, 1). Show that B =
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter5: Orthogonality
Section5.1: Orthogonality In Rn
Problem 28EQ
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![Let n be an integer. Let T = L(R³) be given by the matrix
2 n 1
1
A =
n
n
n
Observe that A is symmetric (hence T is self-adjoint). In the following steps, we
shall find an orthonormal basis of eigenvectors for T. (Note that for convenience of
notation, vectors are written as row vectors.)
a.) Show that vo = (1, -2, 1) is an eigenvector of T with eigenvlaue A = 0.
b.) Let W
=
E(0, T). Let w₁ (1, 0, -1) and w₂
{w₁, w2} is an orthogonal basis of W.
(1, 1, 1). Show that B
n
n
n+1
n+1 n+ 2,
=
g.) Conclude that eo =
e1
c.) Verify that Tw₁ = −2w₂ and Tw2 = (3n)w2 - 3w₁. Let B = [Tw]B. Conclude
that
-3
• (-2 3²).
3n
B =
=
d.) Let X₁ and 2 be the roots of the characteristic equation of B. Show that
(-3, A₁) is an eigenvector of B with eigenvalue λ₁. Similarly, (−3, λ₂) is an
eigenvector of B with eigenvalue X₂
e.) Conclude that v₁
=
=
(−3+ A1, A1, 3 + λ₁) spans E(X₁, T). And that v₂
A2, A2, 3 + A₂) spans E(X₂, T).
1 -2 1
√6' √6' √6),
1
(−3+ A1, A₁, 3+₁), e₂ =
3√√/4+nλ₁
is an orthonormal basis of eigenvectors of T.
==
f.) Show that ||v₁||² = 9||w₁||² + |A1|²||w2||² = 18 + 3(A₁)² = 36 + 9nX₁. Similarly,
||v₂||² = 36 +9nX₂.
1
3√4+nλ₂
(−3+
(−3+ A₂, A₂, 3+ A₂)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F979d0aba-5428-414f-a3ba-5510f0301082%2Ff8dc8fec-9c2a-4b62-90e1-e4d0486baf0c%2Fm9187ks_processed.png&w=3840&q=75)
Transcribed Image Text:Let n be an integer. Let T = L(R³) be given by the matrix
2 n 1
1
A =
n
n
n
Observe that A is symmetric (hence T is self-adjoint). In the following steps, we
shall find an orthonormal basis of eigenvectors for T. (Note that for convenience of
notation, vectors are written as row vectors.)
a.) Show that vo = (1, -2, 1) is an eigenvector of T with eigenvlaue A = 0.
b.) Let W
=
E(0, T). Let w₁ (1, 0, -1) and w₂
{w₁, w2} is an orthogonal basis of W.
(1, 1, 1). Show that B
n
n
n+1
n+1 n+ 2,
=
g.) Conclude that eo =
e1
c.) Verify that Tw₁ = −2w₂ and Tw2 = (3n)w2 - 3w₁. Let B = [Tw]B. Conclude
that
-3
• (-2 3²).
3n
B =
=
d.) Let X₁ and 2 be the roots of the characteristic equation of B. Show that
(-3, A₁) is an eigenvector of B with eigenvalue λ₁. Similarly, (−3, λ₂) is an
eigenvector of B with eigenvalue X₂
e.) Conclude that v₁
=
=
(−3+ A1, A1, 3 + λ₁) spans E(X₁, T). And that v₂
A2, A2, 3 + A₂) spans E(X₂, T).
1 -2 1
√6' √6' √6),
1
(−3+ A1, A₁, 3+₁), e₂ =
3√√/4+nλ₁
is an orthonormal basis of eigenvectors of T.
==
f.) Show that ||v₁||² = 9||w₁||² + |A1|²||w2||² = 18 + 3(A₁)² = 36 + 9nX₁. Similarly,
||v₂||² = 36 +9nX₂.
1
3√4+nλ₂
(−3+
(−3+ A₂, A₂, 3+ A₂)
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