Unit 06 · Algebra I

Change of Basis & Similarity

The same linear map looks like different matrices in different bases. Change of basis is the translation rule, and similarity is the equivalence it induces — the key to hunting for a basis in which the map is simple, which is the whole story of Unit 10.

By the end you can

Compute coordinate vectors and the change-of-coordinates matrix.
Derive and use $[T]_{\beta'}=P^{-1}[T]_{\beta}\,P$.
Prove similarity is an equivalence relation and list its invariants.
Show two matrices are (or cannot be) similar.

Why it matters & where it comes from

Why it exists

A clumsy matrix can become diagonal in the right basis. Change of basis is how we hunt for that better viewpoint, and similarity tells us which matrices represent the same map.

In the real world

Picking a better basis is the whole idea behind PCA, Fourier methods, and decoupling systems of differential equations in engineering and finance.

Story & open questions

The search for the "best" basis for a given map leads straight to eigenvalues and the diagonalization problem of Unit 10 — and, when diagonalization fails, to the Jordan form studied in MAT247.

Coordinates and the change-of-coordinates matrix

Definition

For an ordered basis $\beta=\{v_1,\dots,v_n\}$, the coordinate vector $[v]_\beta$ lists the unique coefficients with $v=\sum a_iv_i$. If $\beta'$ is a second basis, the change-of-coordinates matrix $P$ has the $\beta$-coordinates of the $\beta'$-vectors as its columns, and $$[v]_{\beta}=P\,[v]_{\beta'}.$$ $P$ is invertible, and $P^{-1}$ translates in the other direction.

Example

In $\mathbb{R}^2$ with $\beta'=\{(1,1),(1,-1)\}$ and $\beta$ standard: $P=\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)$. For $v=(5,1)$: solve $a(1,1)+b(1,-1)=(5,1)$, so $a+b=5$, $a-b=1$, giving $[v]_{\beta'}=(3,2)$ — and indeed $P\,(3,2)=(5,1)$. ✓

What happens to the matrix of an operator

Theorem

For an operator $T:V\to V$ and bases $\beta,\beta'$ with change-of-coordinates matrix $P$ (from $\beta'$ to $\beta$), $$[T]_{\beta'}=P^{-1}\,[T]_{\beta}\,P.$$

Proof — just follow the coordinates

Take $\beta'$-coordinates of $v$, translate to $\beta$ ($P$), apply the map ($[T]_\beta$), translate back ($P^{-1}$): $[T(v)]_{\beta'}=P^{-1}[T]_\beta P\,[v]_{\beta'}$.

The matrix that does this for all $v$ is unique, so it is $[T]_{\beta'}$. ∎

Example A map made diagonal

$T(x,y)=(3x+y,\ x+3y)$ has standard matrix $A=\left(\begin{smallmatrix}3&1\\1&3\end{smallmatrix}\right)$.

$T(1,1)=(4,4)=4\,(1,1)$ and $T(1,-1)=(2,-2)=2\,(1,-1)$.

So in the basis $\{(1,1),(1,-1)\}$ the matrix is $\left(\begin{smallmatrix}4&0\\0&2\end{smallmatrix}\right)$ — the same map, now decoupled. This is a preview of eigenvectors (Unit 10).

Similarity

Definition

$A,B\in M_{n\times n}(F)$ are similar if $B=P^{-1}AP$ for some invertible $P$. By the theorem above, similar matrices are exactly the matrices of one operator in different bases.

Theorem Invariants

Similar matrices have the same determinant, trace, rank, characteristic polynomial, and eigenvalues. Quick proofs: $\det(P^{-1}AP)=\det(P)^{-1}\det A\det P=\det A$, and $\operatorname{tr}(P^{-1}(AP))=\operatorname{tr}((AP)P^{-1}) =\operatorname{tr}A$ using $\operatorname{tr}(XY)=\operatorname{tr}(YX)$.

Invariants refute, they do not prove Different trace ⇒ not similar: instant. But equal invariants do not force similarity — $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ and the identity share det, trace, rank, and characteristic polynomial, yet only the identity is similar to the identity ($P^{-1}IP=I$). Deciding similarity in general needs canonical forms (MAT247).

Common pitfalls

• Mind the direction: $[T]_{\beta'}=P^{-1}[T]_{\beta}P$ where $P$ carries $\beta'$-coordinates to $\beta$-coordinates.
• The columns of $P$ are the new basis vectors written in the old coordinates.
• Sharing det/trace/eigenvalues does not make matrices similar — see the checkpoint.
• Similarity uses one basis change on both sides ($P$ and $P^{-1}$); using two unrelated matrices $Q^{-1}AP$ is a different, weaker equivalence.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Invariants

Are similar matrices guaranteed to have the same eigenvalues?

Answer

Yes — they share the characteristic polynomial: $\det(P^{-1}AP-tI)=\det(P^{-1}(A-tI)P)=\det(A-tI)$.

Practice 2 · Determinant

If $B=P^{-1}AP$, prove $\det B=\det A$.

Solution

Multiplicativity: $\det B=\det(P^{-1})\det A\det P=\dfrac{1}{\det P}\det A\det P =\det A$.

Practice 3 · Rigid identity

Which matrices are similar to the identity?

Answer

Only $I$ itself: $P^{-1}IP=I$ for every invertible $P$. (Same for any scalar matrix $cI$.)

Practice 4 · Coordinates

Find $[v]_{\beta'}$ for $v=(5,1)$ and $\beta'=\{(1,1),(1,-1)\}$.

Solution

$a+b=5$ and $a-b=1$ give $a=3$, $b=2$: $[v]_{\beta'}=(3,2)$.

Practice 5 · Diagonalize by hand

Compute the matrix of $T(x,y)=(3x+y,\ x+3y)$ in the basis $\{(1,1),(1,-1)\}$.

Solution

$T(1,1)=4(1,1)$ and $T(1,-1)=2(1,-1)$, so the matrix is $\left(\begin{smallmatrix}4&0\\0&2\end{smallmatrix}\right)$.

Practice 6 · Trace

Using $\operatorname{tr}(XY)=\operatorname{tr}(YX)$, prove similar matrices have equal trace.

Solution

$\operatorname{tr}(P^{-1}(AP))=\operatorname{tr}((AP)P^{-1}) =\operatorname{tr}(A)$.

Practice 7 · Equivalence relation

Prove similarity is reflexive, symmetric, and transitive.

Solution

$A=I^{-1}AI$; if $B=P^{-1}AP$ then $A=(P^{-1})^{-1}B\,P^{-1}$; and $C=Q^{-1}BQ$ gives $C=(PQ)^{-1}A(PQ)$.

Practice 8 · Not similar

Show $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ and $I$ are not similar, even though they share determinant, trace, rank, and characteristic polynomial.

Solution

Everything similar to $I$ equals $I$ (Practice 3), and this matrix is not $I$. Moral: the common invariants are necessary, never sufficient.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT240H1 topic map. It should cover the following items before you move on:

Change of basis. Convert coordinate vectors between ordered bases using transition matrices.
Similarity. Recognize matrices of the same operator in different bases as similar matrices.
Invariants. Know trace, determinant, rank, characteristic polynomial, and eigenvalues are preserved under similarity.
Coordinate discipline. Distinguish vectors from coordinate columns and operators from their representing matrices.

Textbook/source alignment

Use the current UofT MAT240H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The spine is: coordinates depend on basis, vectors do not; changing basis conjugates operator matrices by an invertible transition matrix.

Study routine. Whenever a formula has $P^{-1}AP$, label which basis each matrix converts from and to.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Transition matrix

What are the columns of the change-of-basis matrix from basis $B$ to the standard basis?

Guide

They are the vectors of $B$ written in standard coordinates.

Extension 2 · Similarity

Define what it means for matrices $A$ and $B$ to be similar.

Guide

They are similar if there is an invertible matrix $P$ such that $B=P^{-1}AP$.

Extension 3 · Invariant

Prove determinant is preserved under similarity.

Guide

det$(P^{-1}AP)=\det(P^{-1})\det(A)\det(P)=\det(A)$.

Extension 4 · Concept

Why does similarity preserve eigenvalues?

Guide

Similar matrices represent the same operator in different bases, and algebraically they have the same characteristic polynomial.

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