Equivalent & Similar Matrices

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# Equivalent & Similar Matrices ## 🎯 Learning Objectives After this topic you will be able to: - Define equivalent matrices and similar matrices - Distinguish between equivalence and similarity - Find the relationship between matrices representing the same transformation - Compute similar matrices via $P^{-1}AP$ -...

Equivalent & Similar Matrices

🎯 Learning Objectives

After this topic you will be able to:

Define equivalent matrices and similar matrices
Distinguish between equivalence and similarity
Find the relationship between matrices representing the same transformation
Compute similar matrices via $P^{-1}AP$
Understand that similar matrices share eigenvalues, determinant, trace, and rank

📋 Prerequisites

Matrix Representation (Week 6) — how matrices represent linear transformations
Change of Basis (Week 6) — transition matrices
Matrix Multiplication (Week 1) — necessary for $P^{-1}AP$

1. Intuition: Same Map, Different Coordinates

When you change the coordinate system, the numbers describing a point change — but the point itself doesn't. Similarly, when you change bases, the matrix representing a linear transformation changes — but the transformation itself is the same. (Diagram)

Equivalent matrices: $A$ and $B$ represent the same $T: V \to W$ with different bases in domain and codomain
Similar matrices: $A$ and $B$ represent the same $T: V \to V$ with different bases in the domain (same basis change for input and output)

2. Equivalent Matrices

2.1 Definition

Definition (Equivalent Matrices). Two $m \times n$ matrices $A$ and $B$ are equivalent if there exist invertible matrices $P$ ( $m \times m$ ) and $Q$ ( $n \times n$ ) such that:

> B = PAQ >

Why this matters:

A

represents

T: V \to W

in bases

B, C

B

represents the same

T

in bases

B', C'

. Then

P

is the change-of-basis in

W

and

Q

is the change-of-basis in

V

B = [T]_{C' \leftarrow B'} = P_{C' \leftarrow C} \; [T]_{C \leftarrow B} \; P_{B \leftarrow B'}

2.2 Properties

Property	Explanation
Rank preserved	$\text{rank}(A) = \text{rank}(B)$
Dimensions same	Both are $m \times n$
Row/column operations	$PAQ$ means row ops on $A$ ( $P$ ) and column ops on $A$ ( $Q$ )
Equivalence is...	Reflexive, symmetric, transitive (an equivalence relation)

Key theorem: Two

m \times n

matrices are equivalent iff they have the same rank.

Example 1: Equivalent matrices

> A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} >

and

> B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} >

have rank 1, so they're equivalent.

More interesting:

> A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} >

has rank 2. Any $2 \times 2$ matrix with rank 2 is equivalent to $A$ (and to $I_2$ ).

3. Similar Matrices

3.1 Definition

Definition (Similar Matrices). Two $n \times n$ matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that:

> B = P^{-1}AP >

Why this matters:

A

and

B

represent the same linear operator

T: V \to V

in different bases.

P

is the change-of-basis matrix.

3.2 Properties of Similar Matrices

Similar matrices share many properties:

Property	Why?
Same determinant	$\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A)$
Same trace	$\text{tr}(P^{-1}AP) = \text{tr}(A)$
Same rank	$\text{rank}(P^{-1}AP) = \text{rank}(A)$ ( $P$ is invertible)
Same eigenvalues	$\det(P^{-1}AP - \lambda I) = \det(P^{-1}(A-\lambda I)P) = \det(A-\lambda I)$
Same characteristic polynomial	$\det(P^{-1}AP - \lambda I) = \det(A - \lambda I)$
Same nullity	$\dim(N(P^{-1}AP)) = \dim(N(A))$

What's NOT preserved:

The eigenvectors change (they transform by $P^{-1}$ )
The matrix entries themselves

Example 2: Similar matrices

> A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} >

,

> P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} >

, then

> P^{-1} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} >

.

> B = P^{-1}AP = \begin{bmatrix} 5 & -1 \\ -2 & 0 \end{bmatrix} >

(as computed in Week 6).

Check: $\det(A) = -2$ , $\det(B) = 0 - (-2)(-1) = -2$ ✓ $\text{tr}(A) = 5$ , $\text{tr}(B) = 5$ ✓

4. Equivalence vs. Similarity

Aspect	Equivalence	Similarity
Form	$B = PAQ$	$B = P^{-1}AP$
Same transformation?	$T: V \to W$	$T: V \to V$ (operator)
Bases	Different in domain & codomain	Same basis in domain & codomain
Matrix size	$m \times n$ (need not be square)	Square $n \times n$
Key invariant	Rank	Rank, det, trace, eigenvalues, char poly

Example 3: Distinguishing equivalence and similarity

> A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} >

and

> B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} >

are equivalent (both have rank 1).

Are they similar? Check trace: $\text{tr}(A) = 1$ , $\text{tr}(B) = 1$ . Check determinant: $\det(A) = 0$ , $\det(B) = 0$ . Check eigenvalues: both have eigenvalues 0 and 1.

They could be similar. Find $P$ such that $B = P^{-1}AP$ :

> P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} >

(swap basis vectors).

> P^{-1}AP = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = B >

✓. Yes, they are similar.

5. Edge Cases & Gotchas

Situation	What Happens
$A$ and $B$ are equivalent but not similar	$A$ is $m \times n$ , $m \neq n$ , or they have different trace/det
$A$ is diagonalisable	$A$ is similar to a diagonal matrix
Identity matrix	Only similar to itself (check: $P^{-1}IP = I$ )
Zero matrix	Only similar to itself

6. Common Pitfalls

❌ Pitfall 1: Confusing $P^{-1}AP$ with $PAP^{-1}$

The order matters.

P^{-1}AP

changes basis for vectors, then applies

A

, then changes back.

PAP^{-1}

does the opposite.

❌ Pitfall 2: Thinking "similar" means "equivalent"

Every similar pair is equivalent, but not vice versa (similar requires square matrices and preserves more properties).

❌ Pitfall 3: Forgetting $P$ must be invertible

P

isn't invertible,

B = P^{-1}AP

isn't defined (and

PAP^{-1}

doesn't capture the same transformation in a different basis).

7. Formula Summary Table

Concept	Formula	Invariants
Equivalence	$B = PAQ$ , $P, Q$ invertible	Rank
Similarity	$B = P^{-1}AP$ , $P$ invertible	Rank, det, trace, eigenvalues, char poly
Change of basis for $T: V \to W$	$[T]_{C' \leftarrow B'} = P_{C' \leftarrow C} [T]_{C \leftarrow B} P_{B \leftarrow B'}$	—
Change of basis for $T: V \to V$	$[T]_{B'} = P^{-1} [T]_B P$	—

8. 📝 Practice Questions

Q1: Are these similar?

> A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

,

> B = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix} >

.

Solution: $\text{tr}(A) = 5$ , $\text{tr}(B) = 5$ . $\det(A) = 0$ , $\det(B) = 0$ . Eigenvalues of $A$ : $\det(A - \lambda I) = (1-\lambda)(4-\lambda) - 4 = \lambda^2 - 5\lambda = \lambda(\lambda-5)$ . So eigenvalues are 0 and 5. Eigenvalues of $B$ : 0 and 5.

Yes, they could be similar (same eigenvalues). Q2: Similarity preserves determinant

Prove: If $B = P^{-1}AP$ , then $\det(B) = \det(A)$ .

Proof: $\det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \frac{1}{\det(P)}\det(A)\det(P) = \det(A)$ . ✓ Q3: Are these equivalent?

> A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} >

,

> B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} >

.

Solution: Both have rank 1 and are $2 \times 3$ . Yes, they are equivalent. Q4: Not similar — different trace

> A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} >

,

> B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} >

.

Solution: $\text{tr}(A) = 2$ , $\text{tr}(B) = 3$ . Not similar. Q5: Finding similar matrix

Find $B$ similar to

> A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} >

using

> P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} >

.

Solution:

> P^{-1} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} >

.

> B = P^{-1}AP = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} >

> = \begin{bmatrix} 1.5 & 1.5 \\ 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} >

.

$B$ is diagonal. $\det(B) = 3 = \det(A)$ ✓, $\text{tr}(B) = 4 = \text{tr}(A)$ ✓. Q6: Matrix equivalence classes

How many equivalence classes of $2 \times 3$ matrices are there?

Solution: Equivalence is determined by rank. Rank can be 0, 1, or 2 (min(2,3) = 2). So 3 equivalence classes. Q7: Similarity preserves eigenvalues

Show that if $\lambda$ is an eigenvalue of $A$ , then $\lambda$ is also an eigenvalue of $B = P^{-1}AP$ .

Proof: If $A\mathbf{v} = \lambda\mathbf{v}$ , then $B(P^{-1}\mathbf{v}) = P^{-1}AP(P^{-1}\mathbf{v}) = P^{-1}A\mathbf{v} = P^{-1}(\lambda\mathbf{v}) = \lambda(P^{-1}\mathbf{v})$ . So $P^{-1}\mathbf{v}$ is an eigenvector of $B$ with eigenvalue $\lambda$ . ✓ Q8: Are $A$ and $A^T$ similar?

Is $A$ always similar to $A^T$ ?

Solution: Yes! Every matrix is similar to its transpose. (This is a known theorem, though the proof uses Jordan canonical form.) Q9: Determining similarity by invariants

> A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} >

,

> B = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} >

. Are they similar?

Solution: $\det(A) = 1-4 = -3$ , $\det(B) = -3$ ✓. $\text{tr}(A) = 2$ , $\text{tr}(B) = 2$ ✓. Eigenvalues: $A$ has $\lambda = 3, -1$ (since $(1-\lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = (\lambda-3)(\lambda+1)$ ). $B$ has eigenvalues 3, -1. Similar ✓. Q10: Non-square case

Can a $2 \times 3$ matrix be similar to a $3 \times 2$ matrix?

Solution: No. Similarity is only defined for square matrices. Non-square matrices can be equivalent (if same rank) but not similar.

🔗 Cross-References

Next topic: Affine Subspaces
Week 12 (Eigenvalues): Diagonalisation is finding a similarity transformation to a diagonal matrix
Week 8 (Orthogonal Transformations): Orthogonal similarity $Q^T A Q$ where $Q^T = Q^{-1}$
BSCS2004 (ML Foundations): PCA uses similarity to diagonalise the covariance matrix Join Discord Previous6.2 Matrix RepresentationNext7.2 Affine Subspaces

What this note is really saying

Equivalent & Similar Matrices

🎯 Learning Objectives

📋 Prerequisites

1. Intuition: Same Map, Different Coordinates

2. Equivalent Matrices

2.1 Definition

2.2 Properties

3. Similar Matrices

3.1 Definition

3.2 Properties of Similar Matrices

4. Equivalence vs. Similarity

5. Edge Cases & Gotchas

6. Common Pitfalls

❌ Pitfall 1: Confusing P−1APP^{-1}APP−1AP with PAP−1PAP^{-1}PAP−1

❌ Pitfall 2: Thinking "similar" means "equivalent"

❌ Pitfall 3: Forgetting PPP must be invertible

7. Formula Summary Table

8. 📝 Practice Questions

🔗 Cross-References

Intelligence Hub

❌ Pitfall 1: Confusing $P^{-1}AP$ with $PAP^{-1}$

❌ Pitfall 3: Forgetting $P$ must be invertible