rheisf

2021-12-18

Can someone show me step-by-step how to diagonalize this matrix? Im

William Appel

First step: Find the eigenvalues of your matrix.
Eigenvectors are vectors x such that upon being multiplied by a matrix A, they are only scaled by a number. That is $Ax=\lambda x$, where $\lambda$ is just a number, called the eigenvalue associated with the eigenvector x.
The way to do this is to subtract the $\lambda x$ from both sides to get $Ax-\lambda x=0$. Now factor out the x to get $\left(A-\lambda I\right)x=0$, where I is the identity matrix - note: we need the identity matrix because adding a matrix and a scalar is undefined.
This equation, $\left(A-\lambda I\right)x=0$ has a nontrivial solution (a solution where $x\ne 0$) if and only if $det\left(A-\lambda I\right)=0$ (can you prove this?).
So lets

lenkiklisg7

This equation, $\left(A-\lambda I\right)x=0$ has a nontrivial solution (a solution where $x\ne 0$) if and only if $det\left(A-\lambda I\right)=0$"
We first assume that $B=A-\lambda I$ is invertible (or $det\left(B\right)\ne 0$). If this is true, the equation can be rewritten as:
${B}^{-1}Bx={B}^{-1}0=0$
Since ${B}^{-1}B=I$ (the identify matrix), then $x=0$, which is false. Then B is not invertible or $det\left(B\right)=0$.

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