Linear transformations, matrix representation, change of basis

In one sentence

A matrix is a linear map plus a choice of basis, similarity is what changing that choice does, and the invariants of similarity are exactly the things a question can legitimately ask about a map.

Why the exam asks it

Because "same characteristic polynomial" feels like it ought to mean "same matrix up to basis", and it does not — not even with the minimal polynomial added, once the size reaches $4$. The gap between the invariants you can compute and the invariants that actually classify is where this topic lives, and it is the doorway to canonical forms.

The idea

Given bases $B$ of $V$ and $C$ of $W$, a linear $T : V \to W$ becomes a matrix $[T]_{B}^{C}$: its $j$-th column is the coordinate vector of $T(b_j)$ in the basis $C$. Composition becomes matrix multiplication, which is the reason matrix multiplication is defined the way it is.

Change of basis

For $T : V \to V$ with two bases $B$ and $B'$, and $P$ the change-of-basis matrix whose columns are the new basis vectors written in the old basis,

$$[T]_{B'} = P^{-1}\,[T]_{B}\,P .$$

Two matrices related this way are similar. Getting $P$ and $P^{-1}$ the right way round is a standing source of lost marks; the reliable check is that $P$ takes new coordinates to old, so it must be undone on the far side.

Similar against equivalent

Similar ($B = P^{-1}AP$) is one map in two bases. It preserves rank, trace, determinant, characteristic polynomial, minimal polynomial, eigenvalues with multiplicities, and Jordan form.

Equivalent ($B = PAQ$, with $P, Q$ invertible and unrelated) allows a different basis at each end. It preserves rank and nothing else. Any two matrices of the same size and rank are equivalent — which makes equivalence far too coarse to be interesting, and is exactly why the canonical form under equivalence is the trivial one with an identity block and zeros.

Similar implies equivalent. The converse is wildly false.

What the invariants fail to determine

$$A = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}, \qquad I = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}.$$

Both have characteristic polynomial $(x-1)^2$. They are not similar: $P^{-1}IP = I$ for every $P$, so the identity is similar only to itself, and $A \ne I$. Concretely $A - I$ has rank $1$, so $A$ has a one-dimensional eigenspace where $I$ has a two-dimensional one.

Adding the minimal polynomial helps but does not finish the job. Here it does distinguish them — $A$ has minimal polynomial $(x-1)^2$ and $I$ has $(x-1)$ — but from size $4$ upwards it stops being enough:

$$J_2 \oplus J_2 \quad\text{and}\quad J_2 \oplus J_1 \oplus J_1$$

both have characteristic polynomial $(x-1)^4$ and minimal polynomial $(x-1)^2$, and they are not similar, having different numbers of Jordan blocks. What does classify is the full Jordan structure, which is the next chapter's business.

Trace and determinant

$\operatorname{tr}(AB) = \operatorname{tr}(BA)$ always, even when $AB \ne BA$. This is the reason trace is a similarity invariant: $\operatorname{tr}(P^{-1}AP) = \operatorname{tr}(APP^{-1}) = \operatorname{tr}(A)$.

Two warnings. $\operatorname{tr}(AB) \ne \operatorname{tr}(A)\operatorname{tr}(B)$ — trace is linear, not multiplicative. And the cyclic property is genuinely cyclic: $\operatorname{tr}(ABC) = \operatorname{tr}(BCA)$, but not $\operatorname{tr}(ACB)$.

$\det(AB) = \det(A)\det(B)$, which gives $\det(P^{-1}AP) = \det A$ the same way. Note the contrast: determinant is multiplicative and trace is additive, and options swap them.

AB and BA

For square $A, B$ of the same size, $AB$ and $BA$ have the same characteristic polynomial, hence the same eigenvalues with multiplicities, the same trace and the same determinant. They need not be similar — the standard pair above has $AB \ne 0$ and $BA = 0$, and a nonzero matrix is never similar to the zero matrix.

For non-square $A$ ($m\times n$) and $B$ ($n \times m$), the nonzero eigenvalues of $AB$ and $BA$ still agree, with multiplicities; only the count of zero eigenvalues differs, because the sizes do.

Where intuition breaks

"Same characteristic polynomial, so similar." $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $I$. The most frequently set false option in this topic.

"Same characteristic and minimal polynomial, so similar." True for $n \le 3$, false from $4$ up: $J_2\oplus J_2$ against $J_2\oplus J_1\oplus J_1$. The size restriction is what makes this a good question — the small cases encourage the wrong general belief.

"$\operatorname{tr}(AB) = \operatorname{tr}(A)\operatorname{tr}(B)$." Take $A = B = I_2$: the left side is $2$, the right is $4$. Determinant is the multiplicative one.

"$\operatorname{tr}(ABC) = \operatorname{tr}(ACB)$." Cyclic permutations only. Swapping two of three is not a cyclic permutation.

"$AB$ and $BA$ are similar." Same characteristic polynomial, not similar in general. $AB \ne 0 = BA$ for the standard pair, and nothing nonzero is similar to zero.

"Equivalent matrices have the same eigenvalues." Equivalence preserves only rank. Every rank-$1$ matrix of a given size is equivalent to every other, and their eigenvalues are all over the place.

"$A$ and $A^T$ have different eigenvalues." They have the same characteristic polynomial, since $\det(A^T - xI) = \det((A - xI)^T)$. In fact $A$ and $A^T$ are always similar, though the similarity is not obvious.

"A change of basis is $PAP^{-1}$, not $P^{-1}AP$." Both forms appear, and which is correct depends on how $P$ is defined. Fix the convention — $P$'s columns are the new basis in old coordinates — and derive the rest rather than memorising.

The exam's angle

1. Decide which relation the question is about. Similar, equivalent, or congruent — they have different invariants, and options mix them deliberately.
2. To disprove similarity, find an invariant that differs. Rank, trace, determinant, eigenvalues, rank of $A - \lambda I$, minimal polynomial. Rank of $A - \lambda I$ is the most under-used and often the quickest.
3. To prove similarity, invariants are not enough beyond small sizes. Compare Jordan structures.
4. Use $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ freely — it needs no commutativity and it settles a surprising number of options in one line.
5. Compute $A - I$ or $A - \lambda I$ early. Its rank gives the geometric multiplicity, which is what separates the diagonalisable case from everything else.

The night before

• $[T]_{B'} = P^{-1}[T]_B P$, with $P$'s columns the new basis in old coordinates.
• Similar preserves rank, trace, det, char poly, min poly, eigenvalues, Jordan form.
• Equivalent ($PAQ$) preserves rank and nothing else. Same size and rank $\Rightarrow$ equivalent.
• $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $I$: same char poly, not similar. The identity is similar only to itself.
• Char poly and min poly still fail to classify from $4\times4$ upwards: $J_2\oplus J_2$ against $J_2 \oplus J_1 \oplus J_1$.
• $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ always; $\operatorname{tr}$ is not multiplicative; only cyclic permutations are free.
• $\det(AB) = \det A \det B$.
• $AB$ and $BA$: same char poly, not necessarily similar. Non-square: same nonzero eigenvalues.
• $A$ and $A^T$ are similar.