Chapter 3.5, Problem 21PS

To determine

(a)

The vectors that span the column space of A when A is the sum of two matrices of rank one: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

The column space of A is spanned by vectors u and w.

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Consider the equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

We can clearly see that the vectors u and w span the column space of A.

Thus, the column space of A is spanned by vectors u and w.

(b)

To Calculate:

The vectors that span the row space of A when A is the sum of two matrices of rank one: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$.

The row space of A is spanned by vectors v and z.

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Consider the equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

We can clearly see that the vectors v and z span the row space of A.

Thus, the row space of A is spanned by vectors v and z.

(c) To Fill:

The blank space.

The rank is less than 2 if vectors u and w are collinear/dependent or if vectors v and z are collinear/dependent.

Given:

The rank is less than 2 if ____ or if _____.

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Collinear means that one vector is multiple of another and it is noted that two collinear vectors are always meant to be linearly dependent.

It is observed that u and w are multiples of each other and also v and z are collinear because they are also multiple of each other.

Therefore, the rank of matrix A is less than 2 only when the vectors u and w are dependent or when the vectors v and z are dependent.

(d)

A and its rank if $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

The rank of $A=u{v}^{\text{T}}+w{z}^{\text{T}}$ is $2$

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

Calculation:

Substitute the values of $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$ in equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]\times \left[001\right]+\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\times \left[100\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 000\end{array}\right]+\left[\begin{array}{l}000\\ 000\\ 100\end{array}\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 100\end{array}\right]$

Therefore, the rank of matrix A is $r\left(A\right)=2$

To determine

(b)

To Calculate:

The vectors that span the row space of A when A is the sum of two matrices of rank one: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$.

The row space of A is spanned by vectors v and z.

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Consider the equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

We can clearly see that the vectors v and z span the row space of A.

Thus, the row space of A is spanned by vectors v and z.

(c) To Fill:

The blank space.

The rank is less than 2 if vectors u and w are collinear/dependent or if vectors v and z are collinear/dependent.

Given:

The rank is less than 2 if ____ or if _____.

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Collinear means that one vector is multiple of another and it is noted that two collinear vectors are always meant to be linearly dependent.

It is observed that u and w are multiples of each other and also v and z are collinear because they are also multiple of each other.

Therefore, the rank of matrix A is less than 2 only when the vectors u and w are dependent or when the vectors v and z are dependent.

(d)

A and its rank if $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

The rank of $A=u{v}^{\text{T}}+w{z}^{\text{T}}$ is $2$

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

Calculation:

Substitute the values of $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$ in equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]\times \left[001\right]+\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\times \left[100\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 000\end{array}\right]+\left[\begin{array}{l}000\\ 000\\ 100\end{array}\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 100\end{array}\right]$

Therefore, the rank of matrix A is $r\left(A\right)=2$

To determine

(c) To Fill:

The blank space.

The rank is less than 2 if vectors u and w are collinear/dependent or if vectors v and z are collinear/dependent.

Given:

The rank is less than 2 if ____ or if _____.

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

Calculation:

Collinear means that one vector is multiple of another and it is noted that two collinear vectors are always meant to be linearly dependent.

It is observed that u and w are multiples of each other and also v and z are collinear because they are also multiple of each other.

Therefore, the rank of matrix A is less than 2 only when the vectors u and w are dependent or when the vectors v and z are dependent.

(d)

A and its rank if $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

The rank of $A=u{v}^{\text{T}}+w{z}^{\text{T}}$ is $2$

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

Calculation:

Substitute the values of $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$ in equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]\times \left[001\right]+\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\times \left[100\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 000\end{array}\right]+\left[\begin{array}{l}000\\ 000\\ 100\end{array}\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 100\end{array}\right]$

Therefore, the rank of matrix A is $r\left(A\right)=2$

To determine

(d)

A and its rank if $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

The rank of $A=u{v}^{\text{T}}+w{z}^{\text{T}}$ is $2$

Given:

Sum of two matrices of rank one is A: $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$

Calculation:

Substitute the values of $u=z=\left(1,0,0\right)$and $v=w=\left(0,0,1\right)$ in equation $A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=u{v}^{\text{T}}+w{z}^{\text{T}}$

$A=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]\times \left[001\right]+\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\times \left[100\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 000\end{array}\right]+\left[\begin{array}{l}000\\ 000\\ 100\end{array}\right]$

$A=\left[\begin{array}{l}001\\ 000\\ 100\end{array}\right]$

Therefore, the rank of matrix A is $r\left(A\right)=2$