Home page for accesible maths 4 Linear transformations 4.A The matrix of a linear transformation 4.C Images and kernels

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4.B Eigenvalues and eigenvectors

The reason the matrix in Exercise 4.6 is diagonal is that the new basis vectors are all eigenvectors. Recall the definition:

Definition 4.10:

Given a linear transformation $T:V\to V$ from a vector space $V$ to itself, an eigenvector is a non-zero vector $\vec{0}\neq\vec{x}\in V$ such that $T\vec{x}=\lambda\vec{x}$ for some scalar $\lambda\in F$ , called an eigenvalue.

It is understood that eigenvectors of a square matrix $A$ refer to the eigenvectors of the associated linear transformation $F^{n}\to F^{n}$ , defined by $\vec{x}\mapsto A\vec{x}$ , using the standard basis to write vectors in $F^{n}$ .

In MATH105, techniques were developed to find all eigenvalues and eigenvectors of real square matrices, first by solving the polynomial equation $\det(A-\lambda I_{n})=0$ (for $\lambda\in F$ ), and then for each eigenvalue, finding all eigenvectors by solving a system of linear equations in the coefficients; these techniques still work over arbitrary fields $F$ . Recall that $c_{A}(\lambda):=\det(A-\lambda I_{n})$ is called the characteristic polynomial of $A$ . It is a degree $n$ polynomial with coefficients in $F$ . One of the main benefits of finding eigenvectors is the following:

Theorem 4.11.

If a vector space $V$ has a basis $\mathcal{B}=(\vec{x_{1}},\cdots,\vec{x_{n}})$ consisting of eigenvectors of some linear transformation $T$ , then

{}_{\mathcal{B}}[T]_{\mathcal{B}}=\operatorname{diag}(\lambda_{1},\cdots,% \lambda_{n}),

where $\lambda_{i}$ is the eigenvalue of $\vec{x_{i}}$ .

Proof.

Since $T\vec{x_{i}}=\lambda_{i}\vec{x_{i}}$ , the $i$ th column of the matrix ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ is

[T\vec{x_{i}}]_{\mathcal{B}}=\begin{bmatrix}0&\cdots&\lambda_{i}&\cdots&0\end{% bmatrix}^{T}

, where the $\lambda_{i}$ is in the $i$ th position. So all of the $\lambda_{i}$ ’s are along the diagonal of ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ , with zeros elsewhere. ∎

Exercise 4.12:

Find the eigenvalues, and their corresponding eigenvectors (known as an eigenspace), for each of the following matrices.

i.

$\begin{bmatrix}1&1&1\\ 0&-2&1\\ 0&0&7\end{bmatrix},$
ii.

$\begin{bmatrix}2&1&1\\ 0&1&1\\ 0&0&2\end{bmatrix},$
iii.

$\begin{bmatrix}2&1&-1\\ 0&1&1\\ 0&0&2\end{bmatrix}$

Exercise 4.13:

For each of the matrices in Exercise 4.12, decide whether or not ${\mathbb{R}}^{3}$ has a basis consisting of eigenvectors.

Exercise 4.14:

Let $V:=\operatorname{M}_{{2}}({F})$ be the vector space of $2\times 2$ matrices over a field $F$ . Let $T:V\to V$ be the transpose, defined by $T(A):=A^{T}$ . Then $T$ is a linear transformation. Can you find a basis of $V$ in which $T$ is diagonal?

[End of Exercise]