14.2 Matrix diagonalization

From our understanding of eigenvalues and eigenvectors we have discovered several things about our operator matrix, $A$ . We know that if the eigenvectors of $A$ span $\mathbb{C}^n$ and we know how to express any vector $x$ in terms of $\{v_1, v_2, \dots , v_n\}()$ , then we have the operator $A$ all figured out. If we have $A$ acting on $x$ , then this is equal to $A$ acting on the combinations of eigenvectors. Which we know proves to be fairly easy!

We are still left with two questions that need to be addressed:

1. When do the eigenvectors $\{v_1, v_2, \dots , v_n\}()$ of $A$ span $\mathbb{C}^n$ (assuming $\{v_1, v_2, \dots , v_n\}()$ are linearly independent)?
2. How do we express a given vector $x$ in terms of $\{v_1, v_2, \dots , v_n\}()$ ?

Answer to question #1

Answer to question #2

Aside

Let us recall our knowledge of functions and their basis and examine the role of $V$ . $$x=V\alpha$$ $$\left(\begin{array}{c}{x}_1\\ ⋮\\ x_n\end{array}\right)=V\left(\begin{array}{c}{\alpha }_1\\ ⋮\\ {\alpha }_n\end{array}\right)$$ where $\alpha$ is just $x$ expressed in a different basis : $$x=x_1\left(\begin{array}{c}1\\ 0\\ ⋮\\ 0\end{array}\right)+x_2\left(\begin{array}{c}0\\ 1\\ ⋮\\ 0\end{array}\right)+\dots +x_n\left(\begin{array}{c}0\\ 0\\ ⋮\\ 1\end{array}\right)$$ $$x={\alpha }_1\left(\begin{array}{c}⋮\\ v_1\\ ⋮\end{array}\right)+{\alpha }_2\left(\begin{array}{c}⋮\\ v_2\\ ⋮\end{array}\right)+\dots +{\alpha }_n\left(\begin{array}{c}⋮\\ v_n\\ ⋮\end{array}\right)$$ $V$ transforms $x$ from the standard basis to the basis $\{v_1, v_2, \dots , v_n\}()$

Matrix diagonalization and output

We can also use the vectors $\{v_1, v_2, \dots , v_n\}()$ to represent the output, $b$ , of a system: $$b=Ax=A({\alpha }_1{v}_1+{\alpha }_2{v}_2+\dots +{\alpha }_n{v}_n)$$ $$Ax={\alpha }_1{\lambda }_1{v}_1+{\alpha }_2{\lambda }_2{v}_2+\dots +{\alpha }_n{\lambda }_n{v}_n=b$$ $$Ax=\begin{pmatrix}⋮ & ⋮ & & ⋮\\ v_1 & v_2 & \dots & v_n\\ ⋮ & ⋮ & & ⋮\\ \end{pmatrix}()\left(\begin{array}{c}{\lambda }_1{\alpha }_1\\ ⋮\\ {\lambda }_1{\alpha }_n\end{array}\right)$$ $$Ax=V\Lambda \alpha$$ $$Ax=V\Lambda V^{-1}x$$ where $\Lambda$ is the matrix with the eigenvalues down the diagonal: $$\Lambda =\begin{pmatrix}{\lambda }_1 & 0 & \dots & 0\\ 0 & {\lambda }_2 & \dots & 0\\ ⋮ & ⋮ & \ddots & ⋮\\ 0 & 0 & \dots & {\lambda }_n\\ \end{pmatrix}$$ Finally, we can cancel out the $x$ and are left with a final equation for $A$ : $$A=V\Lambda V^{-1}$$

Interpretation

For our interpretation, recall our key formulas: $$\alpha =V^{-1}x$$ $$b=\sum {\alpha }_i{\lambda }_i{v}_i$$ We can interpret operating on $x$ with $A$ as: $$\left(\begin{array}{c}{x}_1\\ ⋮\\ x_n\end{array}\right)\to \left(\begin{array}{c}{\alpha }_1\\ ⋮\\ {\alpha }_n\end{array}\right)\to \left(\begin{array}{c}{\lambda }_1{\alpha }_1\\ ⋮\\ {\lambda }_1{\alpha }_n\end{array}\right)\to \left(\begin{array}{c}{b}_1\\ ⋮\\ b_n\end{array}\right)$$ where the three steps (arrows) in the above illustration represent the following three operations:

1. Transform $x$ using $V^{-1}$ , which yields $\alpha$
2. Multiplication by $\Lambda$
3. Inverse transform using $V$ , which gives us $b$