Matrix representation of linear transformations: Question bank

This bank leans on the parent note Matrix representation of linear transformations and pokes at ideas from Coordinate vectors and bases, Change of basis, Composition of linear maps, Rank and nullity, Kernel and image, and Eigenvalues and diagonalization.

True or false — justify

The matrix

[T]_{C}^{B}

is a property of the linear map

T

alone.

False. It depends on both the chosen bases

B

and

C

; the same

T

has infinitely many matrices, one per pair of bases.

If two matrices represent the same linear map (via different bases), they must be equal.

False. They need not be equal — they are equivalent (or similar, if

B = C

changes together), related through change-of-basis matrices. See Change of basis.

A linear map

T : V \to W

with

dim V = n

dim W = m

always has an

m \times n

matrix.

True. Columns count domain basis vectors (

n

) and rows count codomain coordinates (

m

), so the shape is fixed regardless of which bases you pick.

Knowing

T

on three arbitrary vectors of

R^{3}

always determines

T

completely.

False. Only if those three vectors form a basis (are linearly independent). If they are dependent,

T

is undetermined on directions they miss.

The zero map

T (v) = 0

has the zero matrix in every choice of bases.

True. Every basis vector maps to

0

, whose coordinate vector is all zeros in any codomain basis, so all columns are zero.

Changing only the codomain basis

C

can change the matrix while

T

is unchanged.

True. The columns store

T (b_{j})

in the $C$ -basis, so recoordinatizing the output rewrites every entry even if inputs are untouched.

[T]_{C}^{B}

is the identity matrix, then

T

must be the identity map.

False.

[T]_{C}^{B} = I

only says each

b_{j}

has the same coordinate list as

c_{j}

; with

B \neq = C

this can be a genuine change-of-basis map, not the identity.

The matrix of

T \circ S

[S] [T]

(in the natural reading order).

False. It is

[T \circ S] = [T] [S]

: apply

S

first, so its matrix sits on the right, matching function composition order. See Composition of linear maps.

Listing the outputs

T (b_{j})

as rows gives a valid matrix for

T

False. That produces the transpose

A^{⊤}

, which represents a different map; the formula

[T (v)]_{C} = A [v]_{B}

forces outputs into columns.

m \times n

matrix with more columns than rows (

n > m

) can represent an injective map.

False. Injectivity needs zero kernel, but an

m \times n

matrix with

n > m

has rank

\leq m < n

, so its nullity is positive — some nonzero input maps to zero. See Rank and nullity.

Why questions

Why must we express

T (b_{j})

in the codomain basis rather than the domain basis?

Because

T (b_{j})

lives in

W

, and coordinates only make sense relative to a basis of the space the vector actually inhabits — so we use

C

, a basis of

W

Why does matrix multiplication use the "row times column" rule instead of something simpler?

It is reverse-engineered so that multiplying two matrices equals composing the two maps; the rule is the price paid so

[T \circ S] = [T] [S]

holds. See Composition of linear maps.

Why is a linear map completely fixed by its values on a basis?

Every vector is a unique combination

v = \sum x_{j} b_{j}

, and linearity gives

T (v) = \sum x_{j} T (b_{j})

— so the finite data

{T (b_{j})}

determines

T

everywhere.

Why can the same map look diagonal in one basis and full in another?

A diagonalizing basis aligns with the map's eigenvectors, so each basis vector merely scales — off-diagonal entries vanish. In a generic basis directions get mixed. See Eigenvalues and diagonalization.

Why do we store outputs in columns rather than rows?

So that

A e_{j}

(which selects column

j

) returns exactly

[T (b_{j})]_{C}

, making

[T (v)]_{C} = A [v]_{B}

come out right by construction.

Why is the kernel of

T

the same as the null space of its matrix

[T]_{C}^{B}

A vector

v

satisfies

T (v) = 0

iff its coordinate vector

[v]_{B}

satisfies

A [v]_{B} = 0

, so kernel and null space correspond one-to-one through the coordinate map. See Kernel and image.

Why does the rank of the matrix not change when you switch bases?

Change of basis multiplies by invertible matrices, which preserve dimension of the image; rank measures that dimension, an intrinsic property of

T

. See Rank and nullity.

Edge cases

What is the matrix of

T : V \to {0}

(map into the zero space)?

It is the unique

0 \times n

matrix — zero rows because

dim {0} = 0

— representing the map that annihilates everything.

Can a linear map from a

0

-dimensional space have a matrix?

Yes: an

m \times 0

matrix (no columns). The only vector in a

0

-dimensional domain is

0

, and

T (0) = 0

, consistent with an empty product.

dim V = dim W

but

T

is not invertible, is the matrix still square?

Yes, the matrix is square (

n \times n

) because it counts basis vectors, not invertibility. It is just a singular square matrix (determinant zero).

What does a column of all zeros in

[T]_{C}^{B}

tell you?

The corresponding basis vector

b_{j}

maps to

0

, so

b_{j}

lies in the kernel of

T

— a direct read-off of part of Kernel and image.

What happens to the matrix if you reorder the basis vectors of

B

Its columns get permuted in the same order; the map is unchanged but its bookkeeping table is shuffled — another reminder the matrix is basis-dependent.

What happens to the matrix if you reorder the codomain basis

C

Its rows get permuted, since each column's coordinate list is reindexed to match the new order of

C

For the identity map

id : V \to V

with the same basis

B

in and out, what is the matrix?

The identity matrix

I

: each

b_{j}

maps to itself, whose coordinates in

B

are

e_{j}

, filling the diagonal with ones. Choosing a different output basis instead gives a change-of-basis matrix.

Recall One-line summary of every trap

The matrix is a basis-dependent bookkeeping table: columns catch outputs in the codomain basis, shape comes from dimensions, and intrinsic facts (rank, kernel, invertibility) survive basis changes while the entries themselves do not.

Connections

Change of basis — why entries change but the map does not.
Composition of linear maps — origin of the multiplication order.
Coordinate vectors and bases — the $[v]_{B}$ every trap relies on.
Rank and nullity — basis-invariant facts read off the matrix.
Kernel and image — zero columns and null space.
Eigenvalues and diagonalization — why some bases simplify $[T]$ .

Question bank — Matrix representation of linear transformations