Calculation of Eigenvalues and Eigenvectors

Topics: Algebra - Linear Transformation

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Calculation of Eigenvalues §

(theorem)

Let $T:V\to V$ with $V$ a vector space on $\mathbb{K}$ with finite dimension.

A scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $T$ if and only if $\det (T-\lambda I_V)=0$ (Linear Operator Determinant).

(corollary)

Let $A\in M_{n\times n}(\mathbb{K})$ be the associated matrix of $T$. A scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ (of $T$) if and only if $\det (A-\lambda I)=0$.

Examples §

Example 1: Given the associated matrix

Let $A=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)$. Let’s find the eigenvalues of $A$.

First, we calculate $(A-\lambda I)$:

$$\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1-\lambda & 1\\ 4& 1-\lambda \end{array}\right)$$

Second, we calculate the determinant of this matrix:

$\det (A-\lambda I)=\det \left(\begin{array}{cc}1-\lambda & 1\\ 4& 1-\lambda \end{array}\right)$

• $=(1-\lambda )(1-\lambda )-4=1-2\lambda +{\lambda }^2-4$
• $={\lambda }^2-2\lambda -3$
• $=(\lambda -3)(\lambda +1)$

Third, we make this last expression equal to $0$, to find the eigenvalues:

$$(\lambda -3)(\lambda +1)=0⟹{\lambda }_1=3,{\lambda }_2=-1$$

Finally, with that we get that the eigenvalues are ${\lambda }_1=3$ and ${\lambda }_2=-1$.

Example 2: Given the linear transformation

Let $T:P_2(\mathbb{R})\to P_2(\mathbb{R})$ be defined as follows:

• $T(f(x))=f(x)+x{f}^{\prime }(x)+f^{\prime }(x)$

Let $\beta =\{1,x,x^2\}$ be a basis for $P_2(\mathbb{R})$.

Let’s calculate the eigenvalues of $T$.

First step §

We calculate $[T{]}_{\beta }$.

• $T(1)=1+x(0)+0x=1+0x+0x^2$
• $T(x)=x+x(1)+1=1+2x+0x^2$
• $T(x^2)=x^2+x(2x)+2x=0+2x+3x^2$

With that:

$$[T{]}_{\beta }=\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)$$

Second step §

Let’s calculate $\det (T-\lambda I_V)=\det ([T{]}_{\beta }-\lambda I)$:

$$\det \left(\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)-\left(\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right)\right)$$

$$=\det \left(\left(\begin{array}{ccc}1-\lambda & 1& 0\\ 0& 2-\lambda & 2\\ 0& 0& 3-\lambda \end{array}\right)\right)$$

$$=(1-\lambda )(2-\lambda )(3-\lambda )$$

$$=-(\lambda -1)(\lambda -2)(\lambda -3)$$

Third step §

We make this last expression we got equal to 0:

$$-(\lambda -1)(\lambda -2)(\lambda -3)=0⟹{\lambda }_1=1,{\lambda }_2=2,{\lambda }_3=3$$

Finally, with that we have that the eigenvalues are ${\lambda }_1=1$, ${\lambda }_2=2$ and ${\lambda }_3=3$.

Calculation of Eigenvectors §

(theorem)

Let $T:V\to V$ and let $\lambda$ be an eigenvalue of $T$.

A vector $x\in V$ is an eigenvector of $T$ (that corresponds to $\lambda )$ if and only if $x\ne 0$ and $x\in \ker (T-\lambda I)$.

Example 1

Let $A=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)$

We have that ${\lambda }_1=3$, ${\lambda }_2=-1$ (we calculated these in example 1 for the calculation of eigenvectors).

Eigenvector corresponding to ${\lambda }_1=3$ §

We’ll calculate the eigenvector that corresponds to ${\lambda }_1=3$:

$$B=A-{\lambda }_{1}I=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)-3\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

$$=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)-\left(\begin{array}{cc}3& 0\\ 0& 3\end{array}\right)$$

$$=\left(\begin{array}{cc}-2& 1\\ 4& -2\end{array}\right)$$

Let $x=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\in {\mathbb{R}}^2$ be an eigenvector that corresponds to ${\lambda }_1=3⟺x\ne 0$ and $x\in \ker (L_B)$. That is:

$$\left(\begin{array}{cc}-2& 1\\ 4& -2\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

$$⟹\left(\begin{array}{c}-2x_1+x_2\\ 4x_1-2x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

From that, we have:

• $-2x_1+x_2=0⟹x_2=2x_1$
• $4x_1-2x_2=0⟹2x_2=4x_1⟹x_2=2x_1$

Therefore, the solutions to the system take the form $x_2=2x_1$. We can write that as:

$$\left(\begin{array}{c}{x}_1\\ 2x_1\end{array}\right)=x_1\left(\begin{array}{c}1\\ 2\end{array}\right)$$

And with that, we finally get that the eigenvector that corresponds to ${\lambda }_1$ is $\left(\begin{array}{c}1\\ 2\end{array}\right)$.

Eigenvector corresponding to ${\lambda }_2=-1$ §

We’ll calculate the vector that corresponds to ${\lambda }_2=-1$.

$$B=A-{\lambda }_{2}I=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)-(-1)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

$$=\left(\begin{array}{cc}1& 1\\ 4& 1\end{array}\right)-\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$$

$$=\left(\begin{array}{cc}2& 1\\ 4& 2\end{array}\right)$$

Let $x=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\in {\mathbb{R}}^2$ be an eigenvector that corresponds to ${\lambda }_2=-1$, if and only if $x\ne 0$ and $x\in \ker (L_B)$. That is:

$\left(\begin{array}{cc}2& 1\\ 4& 2\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

$$⟹\left(\begin{array}{c}2x_1+x_2\\ 4x_1+2x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

From that, we have:

• $2x_1+x_2=0⟹x_2=-2x_1$
• $4x_1-2x_2=0⟹2x_2=-4x_1⟹x_2=-2x_1$

Therefore, the solutions to the system take the form $x_2=-2x_1$. We can write that as:

$$\left(\begin{array}{c}{x}_1\\ -2x_1\end{array}\right)=x_1\left(\begin{array}{c}1\\ -2\end{array}\right)$$

And with that, we finally get that the eigenvector that corresponds to ${\lambda }_2$ is $\left(\begin{array}{c}1\\ -2\end{array}\right)$.

Conclusion §

The eigenvectors of the linear transformation whose associated matrix is $A$ are:

• $\left(\begin{array}{c}1\\ -2\end{array}\right)$, corresponding to ${\lambda }_1=3$
• $\left(\begin{array}{c}1\\ 2\end{array}\right)$, corresponding to ${\lambda }_2=-1$

Example 1 - Uses for these eigenvectors

(concerning example 1)

The set that contains only the found eigenvectors is a basis for $V$:

$$\left\{\left(\begin{array}{c}1\\ -2\end{array}\right),\left(\begin{array}{c}1\\ 2\end{array}\right)\right\}$$

We can build the $Q$ matrix such that $A=Q^{-1}BQ$ (with $B$ a diagonal matrix; diagonalisation) by using the eigenvectors as columns:

$$Q=\left(\begin{array}{cc}1& 1\\ 2& -2\end{array}\right)$$

No Eigenvalues §

There can be linear operators that, depending on the vector space field, can have eigenvalues or not.

Example

Let $A=\left(\begin{array}{cc}=0& -1\\ 1& -\lambda \end{array}\right)$.

Let’s set $\det (A-\lambda I)=0$ to find the eigenvalues:

$$\det \left(\begin{array}{cc}\lambda & -1\\ 1& -\lambda \end{array}\right)=0$$

$$(-\lambda )(-\lambda )+1=0$$

$${\lambda }^2+1=0$$

Upon solving this quadratic equation, we get that ${\lambda }_{1,2}=\pm i$. Note how these eigenvalues exist in $\mathbb{C}$, but not in $\mathbb{R}$.