Linear Algebra: XIII Similar and Diagonalizable Matrices



Similar Matrices

Let \(A\) and \(B\) be $n$ by $n$ square matrices. We say that \(A\) is similar to \(B\), denoted \(A \sim B\), if there exists an invertible matrix \(P\) called the change of basis matrix such that


$$ B = P^{-1} A P $$

Or equivalently,


$$ A = P^{-1} B P \quad \text{or} \quad AP = PB $$

Eigenvalue Invariance

Suppose that $A \sim B$. Then $A$ and $B$ have the same eigenvalues since:


$$ \det(B - \lambda I) = \det(P^{-1} A P - \lambda I) $$
$$ = \det(P^{-1} A P - P^{-1} \lambda I P) = \det(P^{-1} (A - \lambda I) P) $$
$$ = \det(P^{-1}) \det(A - \lambda I) \det(P) = \det(A - \lambda I) $$

Trace, Determinant, and Rank Invariance

It follows that if $A \sim B$, then $\mathrm{tr}(B) = \mathrm{tr}(A)$ since also by the cyclic property of trace:


$$ \mathrm{tr}(B) = \mathrm{tr}(P^{-1} A P) = \mathrm{tr}(A P^{-1} P) = \mathrm{tr}(A) $$

It also follows that $\mathrm{det}(B) = \mathrm{det}(A)$ since by the determinant of a product property:


$$ \mathrm{det}(B) = \mathrm{det}(P^{-1} A P) = \mathrm{det}(P^{-1}) \ \mathrm{det}(A) \ \mathrm{det}(P) = \mathrm{det}(A) $$

Additionally, $\mathrm{rank}(B) = \mathrm{rank}(A)$ since $P$ is invertible so:


$$ \mathrm{rank}(B) = \mathrm{rank}(P^{-1} A P) = \mathrm{det}(A) $$

Diagonalizable Matrices

Suppose that the eigenvectors of an $n$ by $n$ square matrix $A$ are linearly independent and together span $\mathbb{R}^{n}$. Then we can construct a change of basis matrix $P$ with columns that are eigenvectors of $A$ so that $A \sim D$ where $D$ is an $n$ by $n$ diagonal matrix where $d_{ii} = \lambda_i$ by the eigenvalue definition:


$$ AP = PD $$

Or equivalently,


$$ D = P^{-1} A P $$

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