Examples

Row-swap example

Row-swap example

The $4\times 4$ matrix which swaps rows $2$ and $4$ (of a vector or another matrix) works like this:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right)\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{41}& a_{42}& a_{43}& a_{44}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{21}& a_{22}& a_{23}& a_{24}\end{array}\right).$$

To obtain this row-swap matrix, start with the diagonal matrix with $1$’s along the diagonal, and then swap rows $2$ and $4$.

Projection is not invertible

Projection is not invertible

Consider the projection matrix $\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ which sends a vector $(x,y)$ to the vector $(x,0)$. This matrix acts by zero upon any vector $(0,y)$, sending $(0,y)$ to $(0,0)$. This action cannot be undone! (What specific vector would the inverse action send $(0,0)$ to?)

Inverse matrix to find preimage

Inverse matrix to find preimage

Problem: Define

$$A=\left(\begin{array}{cc}1& 2\\ -2& -3\end{array}\right),\mathbf{v}=\left(\begin{array}{c}1\\ 4\end{array}\right).$$

Find a vector $\mathbf{u}$ with the property that $A\mathbf{u}=\mathbf{v}$, meaning that $A$ sends $\mathbf{u}$ to the given $\mathbf{v}$.

Solution: First we compute $A^{-1}$ using the formula. We have $ad-bc=-3--4=1$, so $A^{-1}=\left(\begin{array}{cc}-3& -2\\ 2& 1\end{array}\right)$. Now compute that $A^{-1}\mathbf{v}=\left(\begin{array}{c}-11\\ 6\end{array}\right)$, and this is our answer for $\mathbf{u}$. Check this answer by computing $A\left(\begin{array}{c}-11\\ 6\end{array}\right)=\left(\begin{array}{c}1\\ 4\end{array}\right)$, as expected.

Finding eigenvectors and eigenvalues

Finding eigenvectors and eigenvalues

Problem: Define $A=\left(\begin{array}{cc}1& 6\\ 5& 2\end{array}\right)$. Find the two eigenvalues and some (any) two corresponding eigenvectors of $A$.

Solution: First write $A_{\lambda }=\left(\begin{array}{cc}1-\lambda & 6\\ 5& 2-\lambda \end{array}\right)$. Then $\det A_{\lambda }=0$ means $(1-\lambda )(2-\lambda )-30=0$. Solve this quadratic equation to obtain ${\lambda }_1=7$ and ${\lambda }_2=-4$. We have found the two eigenvalues.

Next, write out the matrices

$$A_{{\lambda }_1}=\left(\begin{array}{cc}-6& 6\\ 5& -5\end{array}\right),A_{{\lambda }_2}=\left(\begin{array}{cc}5& 6\\ 5& 6\end{array}\right).$$

Our goal is to find ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ such that $A_{{\lambda }_i}{\mathbf{v}}_i=0$. We can guess two solutions: let ${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$ and let ${\mathbf{v}}_2=\left(\begin{array}{c}-6\\ 5\end{array}\right)$. To check that these are indeed eigenvectors of the original $A$ corresponding to ${\lambda }_1=7$ and ${\lambda }_2=-4$, compute:

$$\left(\begin{array}{cc}1& 6\\ 5& 2\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}7\\ 7\end{array}\right)=7\left(\begin{array}{c}1\\ 1\end{array}\right),\left(\begin{array}{cc}1& 6\\ 5& 2\end{array}\right)\left(\begin{array}{c}-6\\ 5\end{array}\right)=\left(\begin{array}{c}24\\ -20\end{array}\right)=-4\left(\begin{array}{c}-6\\ 5\end{array}\right).$$

You could also use the procedure in the gray box “Derivation.” For $A_{{\lambda }_1}$ we have $x=-5/6$, and for $A_{{\lambda }_2}$ we have $x=1$. Then we would get ${\mathbf{v}}_1=\left(\begin{array}{c}-6\\ -6\end{array}\right)$ and ${\mathbf{v}}_2=\left(\begin{array}{c}-6\\ 5\end{array}\right)$. Notice that the first eigenvector is a scalar multiple of ours above: $\left(\begin{array}{c}-6\\ -6\end{array}\right)=-6\left(\begin{array}{c}1\\ 1\end{array}\right)$.

This is true in general: if $A\mathbf{v}=\lambda \mathbf{v}$, then $A(x\mathbf{v})=\lambda (x\mathbf{v})$. In other words, you can scale an eigenvector and the result is still an eigenvector with the same eigenvalue. This is because $Ax=xA$ when $x$ is a scalar, and of course $x\lambda =\lambda x$ when both are scalars. In yet other words, the line $⟨\mathbf{v}⟩$ spanned by an eigenvector $\mathbf{v}$, called an eigenline, is preserved by the matrix: every vector on this line is mapped back to something on this line.