3 Properties of eigenvalues and eigenvectors

There are a number of general properties of eigenvalues and eigenvectors which you should be familiar with. You will be able to use them as a check on some of your calculations.

Property 1: Sum of eigenvalues

For any square matrix $A$ :

sum of eigenvalues $=$ sum of diagonal terms of $A$ (called the trace of $A$ )

Formally, for an $n\times n$ matrix $A$ : ${\sum }_{i=1}^n{\lambda }_i=\text{ trace}\left(A\right)$

(Repeated eigenvalues must be counted according to their multiplicity.)

Thus if ${\lambda }_1=4,{\lambda }_2=4,{\lambda }_3=1$ then ${\sum }_{i=1}^3{\lambda }_i=9$ ).

Property 2: Product of eigenvalues

For any square matrix $A$ :

product of eigenvalues $=$ determinant of $A$

Formally: ${\lambda }_1{\lambda }_2{\lambda }_3\cdots {\lambda }_n={\prod }_{i=1}^n{\lambda }_i=det\; <!--nolimits\; -->\left(A\right)$

The symbol $\prod$ simply denotes multiplication, as $\sum$ denotes summation.

Example 7

Verify Properties 1 and 2 for the $3\times 3$ matrix:

$A=\left[\begin{array}{ccc}\hfill 2& \hfill -1& \hfill 0\\ \hfill -1& \hfill 2& \hfill -1\\ \hfill 0& \hfill -1& \hfill 2\end{array}\right]$

whose eigenvalues were found earlier.

Solution

The three eigenvalues of this matrix are:

${\lambda }_1=2-\sqrt{2},{\lambda }_2=2,{\lambda }_3=2+\sqrt{2}$

Therefore

${\lambda }_1+{\lambda }_2+{\lambda }_3=\left(2-\sqrt{2}\right)+2+\left(2+\sqrt{2}\right)=6=\text{ trace}\left(A\right)$

$\text{ whilst}{\lambda }_1{\lambda }_2{\lambda }_3=\left(2-\sqrt{2}\right)\left(2\right)\left(2+\sqrt{2}\right)=4=det\; <!--nolimits\; -->\left(A\right)$

Property 3: Linear independence of eigenvectors

Eigenvectors of a matrix $A$ corresponding to distinct eigenvalues are linearly independent i.e. one eigenvector cannot be written as a linear sum of the other eigenvectors. The proof of this result is omitted but we illustrate this property with two examples.

We saw earlier that the matrix

$A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

has distinct eigenvalues ${\lambda }_1=1{\lambda }_2=2$ with associated eigenvectors

$X^{\left(1\right)}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]X^{\left(2\right)}=\left[\begin{array}{c}\hfill 0\\ \hfill 1\end{array}\right]$

respectively.

Clearly $X^{\left(1\right)}$ is not a constant multiple of $X^{\left(2\right)}$ and these eigenvectors are linearly independent .

We also saw that the $3\times 3$ matrix

$A=\left[\begin{array}{ccc}\hfill 2& \hfill -1& \hfill 0\\ \hfill -1& \hfill 2& \hfill -1\\ \hfill 0& \hfill -1& \hfill 2\end{array}\right]$

had the following distinct eigenvalues   ${\lambda }_1=2-\sqrt{2},{\lambda }_2=2,{\lambda }_3=2+\sqrt{2}$  with corresponding eigenvectors of the form shown:

$X^{\left(1\right)}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill \sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right],X^{\left(2\right)}=\left[\begin{array}{c}\hfill 1\\ \hfill 0\\ \hfill -1\end{array}\right],X^{\left(3\right)}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right]$

Clearly none of these eigenvectors is a constant multiple of any other. Nor is any one obtainable as a linear combination of the other two. The three eigenvectors are linearly independent .

Property 4: Eigenvalues of diagonal matrices

A $2\times 2$ diagonal matrix $D$ has the form

$D=\left[\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$

The characteristic equation

$\left|D-\lambda I\right|=0$ is $\left|\begin{array}{cc}\hfill a-\lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill d-\lambda \hfill \end{array}\right|=0$

i.e. $\left(a-\lambda \right)\left(d-\lambda \right)=0$

So the eigenvalues are simply the diagonal elements $a$ and $d$ .

Similarly a $3\times 3$ diagonal matrix has the form

$D=\left[\begin{array}{ccc}\hfill a\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill c\hfill \end{array}\right]$

having characteristic equation

$\left|D-\lambda I\right|=\left(a-\lambda \right)\left(b-\lambda \right)\left(c-\lambda \right)=0$

so again the diagonal elements are the eigenvalues.

We can see that a diagonal matrix is a particularly simple matrix to work with. In addition to the eigenvalues being obtainable immediately by inspection it is exceptionally easy to multiply diagonal matrices.

Task!

Obtain the products $D_1{D}_2$ and $D_2{D}_1$ of the diagonal matrices

$D_1=\left[\begin{array}{ccc}\hfill a\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill c\hfill \end{array}\right]D_2=\left[\begin{array}{ccc}\hfill e\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill f\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill g\hfill \end{array}\right]$

Answer

Exercise

If ${\lambda }_1,{\lambda }_2,\dots {\lambda }_n$ are the eigenvalues of a matrix $A,$ prove the following:

1. $A^T$ has eigenvalues ${\lambda }_1,{\lambda }_2,\dots {\lambda }_n$ .
2. If $A$ is upper triangular, then its eigenvalues are exactly the main diagonal entries.
3. The inverse matrix $A^{-1}$ has eigenvalues $\frac{1}{{\lambda }_1},\frac{1}{{\lambda }_2},\dots \frac{1}{{\lambda }_n}$ .
4. The matrix $A-kI$ has eigenvalues ${\lambda }_1-k,{\lambda }_2-k,\dots {\lambda }_n-k$ .
5. (Harder) The matrix $A^2$ has eigenvalues ${\lambda }_1^2,{\lambda }_2^2,\dots {\lambda }_n^2$ .
6. (Harder) The matrix $A^k$ ( $k$ a non-negative integer) has eigenvalues ${\lambda }_1^k,{\lambda }_2^k,\dots {\lambda }_n^k.$

Verify the above results for any $2\times 2$ matrix and any $3\times 3$ matrix found in the previous Exercises on page 13.

N.B. Some of these results are useful in the numerical calculation of eigenvalues which we shall consider later.

Answer

1. Using the property that for any square matrix $A$ , det $\left(A\right)$ = det $\left(A^T\right)$ we see that if

   $det\; <!--nolimits\; -->\left(A-\lambda I\right)=0\text{ then }det\; <!--nolimits\; -->{\left(A-\lambda I\right)}^T=0$

   This immediately tells us that $det\; <!--nolimits\; -->\left(A^T-\lambda I\right)=0$ which shows that $\lambda$ is also an eigenvalue of $A^T$ .

2. Here simply write down a typical upper triangular matrix $U$ which has terms on the leading diagonal $u_{11},u_{22},\dots ,u_{nn}$ and above it. Then construct $\left(U-\lambda I\right)$ . Finally imagine how you would then obtain $det\; <!--nolimits\; -->\left(U-\lambda I\right)=0$ . You should see that the determinant is obtained by multiplying together those terms on the leading diagonal. Here the characteristic equation is:

   $\left(u_{11}-\lambda \right)\left(u_{22}-\lambda \right)\dots \left(u_{nn}-\lambda \right)=0$

   This polynomial has the obvious roots ${\lambda }_1=u_{11},{\lambda }_2=u_{22},\dots ,{\lambda }_n=u_{nn}$ .

3. Here we begin with the usual eigenvalue problem $AX=\lambda X$ . If $A$ has an inverse $A^{-1}$ we can multiply both sides by $A^{-1}$ on the left to give

   $A^{-1}\left(AX\right)=A^{-1}\lambda X\text{ which gives}X=\lambda A^{-1}X$

   or, dividing through by the scalar $\lambda$ we get

   $A^{-1}X=\frac{1}{\lambda }X$ which shows that if $\lambda$ and $X$ are respectively eigenvalue and eigenvector of $A$ then ${\lambda }^{-1}$ and $X$ are respectively eigenvalue and eigenvector of $A^{-1}$ .

   As an example consider $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]$ . This matrix has eigenvalues ${\lambda }_1=-1$ , ${\lambda }_2=5$ with corresponding eigenvectors $X_1=\left[\begin{array}{c}\hfill 1\\ \hfill -1\end{array}\right]$ and $X_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$ . The reader should verify (by direct multiplication) that $A^{-1}=-\frac{1}{5}\left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]$ has eigenvalues $-1$ and $\frac{1}{5}$ with respective eigenvectors $X_1=\left[\begin{array}{c}\hfill 1\\ \hfill -1\end{array}\right]$ and $X_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$ .

4., 5. and 6. are proved in similar way to the proof outlined in 3.