Artin Problems: Section 4.1.

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)


\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


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.

(b) Let U,W be subspaces of a vector space V. Show that

T: U \times W \rightarrow V where T(u,w) = u+w

is a linear transformation.

cT(u,w) = cu+cw = T(cu,cw)


T(u+u',w+w') = u+u'+w+w' = T(u,w)+T(u',w')

therefore transformation is linear.

(c) Express the dimension formula for T in terms of dimensions of subspaces of V.

Think of T as being represented by it’s matrix M which takes vectors in U \times W to vectors in V.

Note that U \times W \subset V and not necessarily U \times W =V

dim(ker(T)) = dim(V) - dim(U \times W)

dim(image(T)) = dim(U \times W)

Where dim(U\times W) = dim(U)+dim(W)


dim(T) = dim(ker(T))+dim(image(T))

dim(T) = dim(V) - dim(U \times W)+dim(U \times W)


dim(T) = dim(V) - dim(U)-dim(W)+dim(U)+dim(W) = dim(V)

This expresses the dimension formula for T in terms of subspaces of V. Noting that V \subset V.


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