Similar Matrices

We say that two $n \times n$ matrices $A$ and $B$ are similar

  • if for some invertible $M$ we can write $B = M^{-1} A \, M$

Invariant Subspaces

Invariant subspace

  • we know that if $x$ is the eigenvector of $A$, then $Ax = \lambda x$
  • $A$ only "streches" $x$ - but it remains in the same 1-dimensional subspace
  • so such subspace $S$ formed by $x$ is invariant:
  • if $x \in S \Rightarrow Ax \in S$

Invariant column space

  • If there exists such $M$ that $AM = MB$
  • then the $\text{range}(A)$ (the Column Space of $A$) is invariant


Suppose $A$ has all its eigenvalues

  • then if we diagonalize $A$, we have $S^{-1} A \, S = \Lambda$
  • so $A$ is similar to $\Lambda$
  • Here $M = \Lambda$
  • we may take another $M \ne \Lambda$ and will get another matrix similar to $A$ (not necessarily diagonal)

A family of similar matrices for $A$ is a set of all matrices similar for $A$ (for different $M$s)

Why Similar?

What is similar about such $A$ and $B$?

  • they have the same eigenvalues!

Let's check it:

  • let $A \mathbf x = \lambda \mathbf x$
  • and $B = M^{-1} A \, M$
  • then $A \mathbf x = A I \mathbf x = A M M^{-1} \mathbf x = \lambda \mathbf x$
  • now let's multiply by $M^{-1}$ on the left:
  • $\underbrace{M^{-1} A \, M}_{B} \, M^{-1} \mathbf x = M^{-1} \lambda \mathbf x$
  • $B M^{-1} \mathbf x = M^{-1} \lambda \mathbf x$
  • Let $\mathbf x^*$ be $M^{-1} \mathbf x$, so we have
  • $B \mathbf x^* = \lambda \mathbf x^*$
  • so matrices $A$ and $B$ share the same eigenvalue $\lambda$
  • but not the eigenvector! $\mathbf x \ne \mathbf x^*$


For diagonalization

  • $A = S^{-1} \Lambda \, S$ eigenvalues stay the same, but eigenvectors become unit vectors

What if for some $i \ne j$, $\lambda_i = \lambda_j$?

  • there might be not enough eigenvectors to span $\mathbb R^n$
  • i.e. columns of $A$ are not linearly independent

Other Things

There are other things that make these matrices similar.

$M$ does not change:

  • eigenvalues $\lambda_i$ (as discussed earlier)
  • Trace and Determinant (because $\text{tr}(A) = \sum \lambda_i$ and $\text{det}(A) = \prod \lambda_i$)
  • rank, and therefore number of independent eigenvectors

$M$ does change

Jordan Form

Suppose $A$ has a family of similar matrices

  • then the Jordan Form is the most diagonal matrix of the family


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