is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Similar Matrices


MathJax TeX Test Page $$If A,B \in \mathbb{M}_n\text{, then A is similiar to B means there exists an invertible matrix}$$ $$P \in \mathbb{M}_n \text{ such that } A = P^{-1}BP$$


(The following are equivalent)

MathJax TeX Test Page $$\text{1. A is similar to B}$$ $$\text{2. There exists an ordered basis } \mathcal{B} \text{ for } \mathbb{R}^n \text{ such that } [A]_{\mathcal{B}} = B$$ This follows from the change of basis formula. $[A]_{\mathcal{E}} = \underset{E \leftarrow B}{P}[A]_{\mathcal{B}}\underset{E \leftarrow B}{P}^{-1}$ $$\text{3. There exists a linear transformation T and ordered bases } \mathcal{B}_1 , \mathcal{B}_2 \\ \text{ such that } A = [T]_{\mathcal{B}_1}, B = [T]_{\mathcal{B}_2}$$ Let $[T] = A$. Let $\mathcal{B}_1 = \mathcal{E}, \mathcal{B}_2 = \mathcal{B}$. We can recreate 2.

Statements and Proofs of Them

MathJax TeX Test Page $$\text{If A is similar to B, } det(A) = det(B)$$ $det(A) = det\left(P^{-1}BP\right)\\$ $det(A) = det(P^{-1})det(P)det(B) = det(B)$ I will add more to this later.

Relation to Eigenvectors and Eigenvalues

MathJax TeX Test Page Let's say we're given a linear transformation in $\mathbb{R}^2$. $$v_1 = \begin{bmatrix}2 \\ 1\end{bmatrix},v_2 = \begin{bmatrix}1 \\ 4\end{bmatrix}$$ $T(v_1) = 2v_1, T(v_2) = 3v_2$. Let's consider $\mathcal{B} = \{v_1, v_2\}$. We want to know what $T_{\mathcal{E}}$ is. $$T_{\mathcal{B}} = \begin{bmatrix}2 & 0 \\0 & 3\end{bmatrix}$$ $$T_{\mathcal{E}} = \underset{\mathcal{E} \leftarrow \mathcal{B}}{P}\begin{bmatrix}2 & 0 \\0 & 3\end{bmatrix}\underset{\mathcal{E} \leftarrow \mathcal{B}}{P}^{-1}$$ $$\underset{\mathcal{E} \leftarrow \mathcal{B}}{P} = \begin{bmatrix}[v_1]_\mathcal{E} & [v_2]_\mathcal{E}\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 1 & 4\end{bmatrix}$$ $$T_{\mathcal{E}} = \frac{1}{7}\begin{bmatrix}2 & 1 \\ 1 & 4\end{bmatrix}\begin{bmatrix}2 & 0 \\0 & 3\end{bmatrix}\begin{bmatrix}4 & -1 \\ -1 & 1\end{bmatrix}$$ Notice that the first matrix has the eigenvalues of T. The second matrix has the corresponding eigenvalues. The process of turning $T_{\mathcal{E}}$ into the right side of the equation is called diagonalization. There are many reasons we like this form. Read this for more.

Quadratic Forms

Change of Basis