Discussion. Recall that once we have obtained the eigenvalues $\lambda$ of a square matrix $\mathit{A}$, we solve the below equation for $\mathit{x}$ to obtain the corresponding eigenvectorslink:

By definitionlink, the set of vectors $\mathit{x}$ that satisfies such a system is the null space of $\mathit{A}-\lambda \mathit{I}$. The eigenspace associated with $\lambda$ is defined to be this null space.

Note that an eigenspace always contains the zero vector because $\mathit{x}=0$ is a solution to $\text{(1)}$. However, an eigenvector cannot be the zero vector by definitionlink. In other words, all the vectors that reside in the eigenspace are eigenvectors of $\lambda$ except for the zero vector.

Solution. The characteristic polynomiallink of $\mathit{A}$ is:

Therefore, the eigenvalues of $\mathit{A}$ are ${\lambda }_1=3$ and ${\lambda }_2=7$. Next, let's find eigenvectors ${\mathit{x}}_1$ and ${\mathit{x}}_2$ corresponding to ${\lambda }_1$ and ${\lambda }_2$ respectively.

Taking the first row:

Let $x_{12}=t$ where $t$ is some scalar. This means that $x_{11}=-3t$. Therefore, an eigenvector corresponding to eigenvalue ${\lambda }_1$ can be expressed as:

The eigenspace ${\mathcal{E}}_1$ associated with ${\lambda }_1$ is the vector space below:

We can think of this eigenspace as a collection of eigenvectors corresponding to ${\lambda }_1$. Again, keep in mind that any vector in this eigenspace is an eigenvector associated with ${\lambda }_1$ except the zero vector.

Next, let's find the eigenvectors corresponding to the eigenvalue ${\lambda }_2=7$. We proceed like so:

Let $x_{22}=t$ where $t\in \mathbb{R}$. This means $x_{21}$ is:

The eigenspace ${\mathcal{E}}_2$ corresponding to eigenvalue ${\lambda }_2$ is:

Proof. By definitionlink, the eigenspace of an eigenvalue $\lambda$ is:

By theorem, the null space of any $m\times n$ matrix is a space of ${\mathbb{R}}^n$. This completes the proof.

Solution. The characteristic polynomial of $\mathit{A}$ is:

Therefore, the eigenvalues of $\mathit{A}$ are ${\lambda }_1=6$ and ${\lambda }_2=-1$.

Let's find an eigenvector associated with ${\lambda }_1=6$ like so:

Here, $x_{11}$ is a basic variablelink while $x_{12}$ is a free variablelink. Let $x_{12}=t$ where $t$ is some scalar. The solution can now be expressed as:

Therefore, the eigenspace ${\mathcal{E}}_1$ associated with ${\lambda }_1$ is:

The basis for ${\mathcal{E}}_1$ is:

Next, let's find the eigenvectors corresponding to ${\lambda }_2=-1$. We proceed like so:

Let $x_{22}=t$ where $t\in \mathbb{R}$. The solution can be expressed as:

The eigenspace ${\mathcal{E}}_2$ is:

The basis for ${\mathcal{E}}_2$ is:

The eigenbasis for ${\mathbb{R}}^2$ is the union of ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$ that is:

Note that the combined basis ${\mathcal{B}}_1\cup {\mathcal{B}}_2$ is guaranteed to be linearly independent given that they correspond to distinct eigenvalues. We will prove this laterlink in this guide.

Solution. The characteristic polynomial of $\mathit{A}$ is:

The eigenvalue of $\mathit{A}$ is thus $\lambda =2$. The corresponding eigenvectors are obtained by:

Here, $x_1$ is a free variable and $x_2$ is a basic variable. Let $x_1=t$ where $t$ is some scalar. The solution set can now be expressed as:

Therefore, the eigenspace $\mathcal{E}$ associated with $\lambda =2$ is:

The basis for $\mathcal{E}$ is:

Since the eigenspace $\mathcal{E}$ is spanned by one basis vector, the dimensionlink of $\mathcal{E}$ is $1$. By definitionlink, the eigenbasis must be a basis for ${\mathbb{R}}^2$ in this case. However, one basis vector cannot span ${\mathbb{R}}^2$, which means there is no eigenbasis for $\mathit{A}$.

Proof. Let ${\lambda }_1$, ${\lambda }_2$, $\cdots$, ${\lambda }_k$ be distinct eigenvalues with corresponding eigenspace ${\mathcal{E}}_1$, ${\mathcal{E}}_2$, $\cdots$, ${\mathcal{E}}_k$ spanned by basis ${\mathcal{B}}_1$, ${\mathcal{B}}_2$, $\cdots$, ${\mathcal{B}}_k$ respectively. Suppose the union of the eigenvectors is:

Where $n\ge k$. Note that $n$ may be greater than $k$ because the dimension of the eigenspace corresponding to an eigenvalue may be greater than $1$.

Now, consider the criterialink for linear independence:

Where $c_1$, $c_2$, $\cdots$, $c_n$ are some scalar coefficients. Our goal is to show that all of these coefficients must equal zero for $\text{(2)}$ to hold - this will imply that $\{{\mathit{v}}_1,{\mathit{v}}_2,\cdots ,{\mathit{v}}_n\}$ is linearly independent by definitionlink. Let's group the eigenbasis vectors with the same eigenvalue:

Here, we've color-coded the eigenbasis vectors with the same eigenvalue - this is only an example to demonstrate what we mean by the grouping of eigenvectors. For instance, in this case, ${\mathit{v}}_1$ and ${\mathit{v}}_2$ are the eigenbasis vectors of ${\mathcal{B}}_1$ corresponding to the first eigenvalue. Let's define the following vectors:

We can rewrite $\text{(3)}$ as:

For $\text{(5)}$ to hold, the following must be true:

To understand why, suppose these eigenvectors are not equal to the zero vector:

We can make ${\mathit{w}}_1$ in $\text{(5)}$ the subject to get:

This means that we can express ${\mathit{w}}_1$ as a linear combination of the other vectors ${\mathit{w}}_2$, ${\mathit{w}}_3$, $\cdots$, ${\mathit{w}}_k$. In other words, ${\mathit{w}}_1$ is linearly dependent on the other vectors. Now, recall that all the eigenvectors corresponding to distinct eigenvalues are linearly independent by theoremlink. This means that ${\mathit{w}}_1$, ${\mathit{w}}_2$, $\cdots$, ${\mathit{w}}_k$ cannot be eigenvectors, which implies they must be zero vectors. This is why $\text{(6)}$ holds.

Gong back to $\text{(4)}$, we have that:

Because basis vectors are linearly independent by definitionlink, the following is true:

Since all the coefficients must be zero for $\text{(3)}$ to hold, the union of the basis $\{{\mathit{v}}_1,{\mathit{v}}_2,\cdots ,{\mathit{v}}_n\}$ is linearly independent by definitionlink. This completes the proof.