Problems from Chapter 4: Linear Operations, Section 1: The Dimension Formula

Problem 4.1.1

Let $A,B$ be respectively $l \times m, n \times p$ matrices. Show that $M \rightarrow AMB$ is a linear transformation $F^{m\times n} \rightarrow F^{l \times p}$

Recall that a linear transformation is a map $T(x)$ such that $T(x+y) = T(x)+T(y)$ and $cT(x) = T(cx)$ where $x,y$ are vectors and $c$ is a scalar constant.

First note that the map is defined only when $M \in F^{m \times n}$ that is when it has m rows and n columns.

Then observe that $cM \rightarrow A(cM)B = c(AMB)$ by linearity of matrices.
Let $N\in F^{m \times n}$ then $M+N \rightarrow A(M+N)B = AMB+ANB$ by distributivity of matrices.

So the map is linear.

A comment that this is really a very trivial problem and just follows from the composition of matrices being a linear map, however the question is an intuition testing question and so requires a more detailed answer.

Problem 4.1.2

Let $v_1,...,v_n$ be elements of a vector space $\mathbb{V}$ . Prove that $\phi: F^n \rightarrow \mathbb{V}$ defined by

$\phi(X): = v_1x_1+...+v_nx_n = \sum v_ix_i$

is a linear transformation.

From the definitions:

$c\phi(X) = c \sum v_ix_i = \sum cv_ix_i = \phi(cX)$

and

$\phi(X+Y) = \sum v_i(x_i+y_i) = \sum (v_ix_i+v_iy_i) = \sum v_ix_i + \sum v_iy_i = \phi(X) + \phi(Y)$

Therefore the map is linear.

Problem 4.1.3

Let $A$ be an $m \times n$ matrix. Use the dimension formula to prove that the space of solutions of the linear system $AX = 0$ has dimension at least $n-m$ .

The dimension formula tells us

$dim(A) = dim(image(A))+dim(Ker(A))$ .

Here the kernel of $A$ are the solutions to $AX = 0$ by definition of the kernel.

A $m \times n$ matrix is a linear map $\mathbb{F}^n \rightarrow \mathbb{F}^m$ . With $dim(A) = n$

Therefore $dim(image(A)) \leq m$ , then by dimension formula

$dim(ker(A)) = dim(A) - dim(image(A)) \geq n-m$

Problem 4.1.4

Prove that every $m \times n$ matrix of rank 1 has the form $A = XY^T$ , where $X,Y$ are $m,n$ dimensional column vectors. How uniquely determined are these vectors?

We need to show that every $XY^T$ has rank one, and that every matrix $A$ has the form $XY^T$ .

Let $X = (x_1,...,x_m)$ and $Y = (y_1,...,y_n)$

Then the $i^{th}$ row of $XY^T$ is

$x_iy_1,...,x_iy_n = x_iY$

Each row is in the span of $Y$ therefore $XY^T$ has rank one. It is useful to think of this as ‘the codomain of $A$ has dimension one’.

Let $A$ be an arbitrary $m \times n$ rank 1 matrix. Then it’s rows are linear combinations of some vector $Y = (y_1,...,y_n)$ . Therefore each row is of the form

$x_iv_1,...,x_iv_n$

If we assign the $x_i$ , into a vector $X = (x_1,...,x_m)$ . Then we get that

$A = XY^T$

This construction depends on a choice of $Y$ , which then determines $X$ . This choice could be any scalar multiple of a known $Y = (y_1,...,y_n)$ . Thus the choice is not very unique at all, it is any real number other than zero.

There is however a canonical choice, which is

$Y: y_1+...+y_n = 1$

that is the ‘normal’ or ‘orthonormal’ element in the span of $Y$ .

Problem 4.1.5

(a) Let $U,W$ be vector spaces over a field $F$ . Show that the two operations

$(u,w)+(u',w') = (u+u',w+w')$ and $c(u,w) = (cu,cw)$

make the product set $U \times W$ into a product vector space.

$U \times W$ contains the zero element as $0 \in U, 0 \in W$ . Closed under addition as $U,W$ closed under addition. Closed under scalar multiplication as $U,W$ closed under scalar multiplication.