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

The eigenvalue $\lambda=5$ repeats twice so the algebraic multiplicity of $\lambda$ is $2$.

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

The eigenvalues of $\boldsymbol{A}$ are $\lambda_1=3$ and $\lambda_2=4$, and they both occur once. Therefore, the algebraic multiplicity of $\lambda_1$ is $1$ and that of $\lambda_2$ is also $1$.

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

Therefore, the eigenvalues of $\boldsymbol{A}$ are $\lambda_1=3$ and $\lambda_2=4$. Let's now find the geometric multiplicity of $\lambda_1$ and $\lambda_2$.

The null spacelink of $\boldsymbol{A}-\lambda_1\boldsymbol{I}$ is the set of vectors $\boldsymbol{x}$ such that $(\boldsymbol{A}-\lambda_1\boldsymbol{I})\boldsymbol{x}=\boldsymbol{0}$. In matrix form, this is:

We have that $x_1$ is a free variablelink and $x_2$ is a basic variablelink. Let $x_1=t$ where $t$ is some scalar. The solution of the homogeneous system is:

Therefore, the basislink for the null space of $\boldsymbol{A}-\lambda_1\boldsymbol{I}$ is:

The nullity is the dimensionlink of the null space of $\boldsymbol{A}-\lambda_1\boldsymbol{I}$. In other words, the nullity is equal to the number of basis vectors for the null space of $\boldsymbol{A}-\lambda_1\boldsymbol{I}$. Since the basis of the null space consists of a single vector, we have that the nullity of $\boldsymbol{A}-\lambda_1\boldsymbol{I}$ is $1$. Therefore, the geometric multiplicity of $\lambda_1$ is $1$.

Similarly, let's now find the geometric multiplicity of $\lambda_2=4$. The matrix form of $\boldsymbol{A}-\lambda_2\boldsymbol{I}$ is:

Here, $x_1$ is a basic variable while $x_2$ is a free variable. Therefore, we let $x_2=t$ where $t\in\mathbb{R}$. The first row gives us:

Therefore, the solution of the homogenous system is:

Therefore, the basis for the null space of $\boldsymbol{A}-\lambda_2\boldsymbol{I}$ is:

Because the nullity of $\boldsymbol{A}-\lambda_2\boldsymbol{I}$ is $1$, the geometric multiplicity of $\lambda_2$ is $1$.

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

The eigenvalue of $\boldsymbol{A}$ is $\lambda=5$. Since $\lambda=5$ repeats twice, we have that the algebraic multiplicity of $\boldsymbol{A}$ is $2$. Let's now find the geometric multiplicity of $\lambda$. The matrix form of $\boldsymbol{A}-\lambda\boldsymbol{I}$ is:

We have that $x_1$ and $x_2$ are both free variables. Let $x_1=r$ and $x_2=t$ where $r$ and $t$ are some scalars. The solution can be expressed as:

The eigenspace $\mathcal{E}$ of $\lambda=5$ is:

The basis $\mathcal{B}$ of $\mathcal{E}$ is:

Since $\mathcal{B}$ has $2$ basis vectors, the geometric multiplicity of $\lambda$ is $2$. Let's summarize the results below:

We will later prove that the algebraic multiplicity is always at least as large as the geometric multiplicity.

Proof. Recall the characteristic equationlink:

By theoremlink, this means that the matrix $\boldsymbol{A}-\lambda\boldsymbol{I}$ is not invertiblelink. By theoremlink, we have that the homogeneous system $(\boldsymbol{A}-\lambda\boldsymbol{I})\boldsymbol{x}=\boldsymbol{0}$ has at least one non-trivial solution $\boldsymbol{x}$. This means that the null space of $\boldsymbol{A}-\lambda\boldsymbol{I}$ contains at least one eigenvector for each eigenvalue $\lambda$.

By theoremlink, we know that scalar multiples of the eigenvector are also eigenvectors. Therefore, the eigenspace contains a set of all linear combinations of at least one eigenvector. We conclude that the eigenspace of every eigenvalue must have a dimension of at least $1$.

This completes the proof.

Proof. Let $\boldsymbol{A}$ be an $n\times{n}$ matrix. By theoremlink, we know that the degree of the characteristic polynomial of $\boldsymbol{A}$ is $n$. This means that an eigenvalue can occur as a root at most $n$ times. In other words, the algebraic multiplicity of an eigenvalue is at most $n$. This completes the proof.

Proof. By theoremlink, if $\boldsymbol{A}$ is an $n\times{n}$ matrix, then the characteristic polynomial of $\boldsymbol{A}$ will have degree $n$. A polynomial with degree $n$ has exactly $n$ roots (including duplicates). Since the roots of the characteristic polynomial correspond to the eigenvalues of $\boldsymbol{A}$, we conclude that the sum of the algebraic multiplicities of the eigenvalues must be equal to $n$. This completes the proof.

Proof. Suppose $\boldsymbol{A}$ has the following $n$ linearly independent eigenvectors:

However, some of these eigenvectors may be associated with the same eigenvalue. For instance, suppose $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are associated with the same eigenvalue $\lambda=4$. Since $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are linearly independent, the number of basis vectors for the eigenspacelink of $\lambda=4$ must be at least $2$. The reason we say "at least" here is that the number of basis vectors for $\lambda=4$ may potentially be greater than $2$. For instance, consider the following case:

The eigenspace for the eigenvalue may be spanned by the following basis vectors:

Here, because there are $3$ basis vectors, the corresponding geometric multiplicity is $3$. Note that we will later show that such a case is impossible.

By repeatedly applying this logic to the rest of the eigenvectors in \eqref{eq:lwwwN9VKQ6EzRnoSxiM}, we conclude that the sum of geometric multiplicities is greater than or equal to $n$. This completes the proof.

Proof. Suppose we have an $n\times{n}$ matrix $\boldsymbol{A}$ and an invertible $n\times{n}$ matrix $\boldsymbol{P}$. By theoremlink, we know that similarlink matrices have the same characteristic polynomial:

Our goal is to derive an expression for the right-hand side of \eqref{eq:IHhrbeAVsRoFdBeXaWV} that includes a term representing the geometric multiplicity of an eigenvalue $\lambda_i$.

To simplify notations, let $k$ denote the geometric multiplicity of eigenvalue $\lambda_i$ and let $\{\boldsymbol{x}_1,\boldsymbol{x}_2, \cdots,\boldsymbol{x}_k\}$ be the basis for the corresponding eigenspace $\mathcal{E}_i$. We define an $n\times{n}$ invertible matrix $\boldsymbol{P}$ whose first $k$ columns are these basis vectors:

Note that there exists such an invertible matrix $\boldsymbol{P}$ because the first $k$ columns are linearly independent by definition of basis vectors. We can ensure $\boldsymbol{P}$ is invertible by coming up with a set of vectors $x_{k+1}$, $x_{k+2}$, $\cdots$, $x_{n}$ that is also linearly independent. By theoremlink, an $n\times{n}$ matrix with $n$ linearly independent columns is invertible. This guarantees the existence of an invertible matrix $\boldsymbol{P}$.

We now express $\boldsymbol{P}$ as a block matrixlink:

Where:

• $\boldsymbol{B}$ is an $n\times{k}$ sub-matrix whose columns are filled with the eigenbasis vectors.

• $\boldsymbol{C}$ is an $n\times(n-k)$ sub-matrix whose columns are $\boldsymbol{x}_{k+1}$, $\boldsymbol{x}_{k+2}$, $\cdots$, $\boldsymbol{x}_{n}$.

Similarly, let's represent the inverse matrix $\boldsymbol{P}^{-1}$ using block format:

Where:

• $\boldsymbol{D}$ is a $k\times{n}$ sub-matrix.

• $\boldsymbol{E}$ is a $(n-k)\times{n}$ sub-matrix.

By theoremlink, the matrix product $\boldsymbol{P}^{-1}\boldsymbol{P}$ can be expressed as:

Where:

• $\boldsymbol{DB}$ is an $k\times{k}$ sub-matrix.

• $\boldsymbol{DC}$ is an $k\times(n-k)$ sub-matrix

• $\boldsymbol{EB}$ is an $(n-k)\times{k}$ sub-matrix.

• $\boldsymbol{EC}$ is an $(n-k)\times(n-k)$ sub-matrix.

By definitionlink, a matrix multiplied by its inverse matrix is equal to the identity matrix, that is, $\boldsymbol{P}^{-1}\boldsymbol{P}=\boldsymbol{I}_n$. Therefore:

• $\boldsymbol{DB}=\boldsymbol{I}_k$.

• $\boldsymbol{EC}=\boldsymbol{I}_{n-k}$.

• $\boldsymbol{DC}$ is an $k\times(n-k)$ sub-matrix whose entries are all zeroes.

• $\boldsymbol{EB}$ is an $(n-k)\times{k}$ sub-matrix whose entries are all zeroes.

Now, consider $\boldsymbol{P}^{-1}\boldsymbol{AP}$. By theoremlink and theoremlink, $\boldsymbol{P}^{-1}\boldsymbol{AP}$ can be expressed as:

Because $\lambda_i$ is an eigenvalue of $\boldsymbol{A}$ and $\boldsymbol{x}_i$ is a corresponding eigenvector, the following equation holds true by definition:

We now multiply both sides by $\boldsymbol{D}$. Note that this is possible because:

• $\boldsymbol{D}$ has shape $k\times{n}$.

• $\boldsymbol{A}$ has shape $n\times{n}$.

• $\boldsymbol{x}_i$ has shape $n\times{1}$.

Some algebraic manipulation yields:

This means that:

Similarly, we have that $\boldsymbol{EA}=\lambda_i\boldsymbol{E}$. Now we go back to \eqref{eq:jBeRUvHtoFRDerjz6Sl} and perform the following substitutions:

Subtracting both sides by $\lambda\boldsymbol{I}$ where $\lambda$ is any eigenvalue of $\boldsymbol{A}$ yields:

We now take the determinant of both sides. By theoremlink, we have that:

Next, by theoremlink, we get:

We substitute this into \eqref{eq:IHhrbeAVsRoFdBeXaWV} to finally get:

Because $(\lambda_i-\lambda)$ appears $k$ times, the algebraic multiplicity of $\lambda_i$ is at least $k$. This completes the proof.

Proof. We will prove $(1)\implies(2)\implies(3)\implies(1)$. We start by assuming $(1)$ that the $n\times{n}$ matrix $\boldsymbol{A}$ is diagonalizable. By theoremlink, since $\boldsymbol{A}$ is diagonalizable, $\boldsymbol{A}$ has $n$ linearly independentlink eigenvectors $\boldsymbol{v}_1$, $\boldsymbol{v}_2$, $\cdots$, $\boldsymbol{v}_n$.

Suppose $\boldsymbol{A}$ has $t$ distinct eigenvalues and let the corresponding geometric multiplicities be $k_1$, $k_2$, $\cdots$, $k_t$ and the algebraic multiplicities be $m_1$, $m_2$, $\cdots$, $m_t$.

By theoremlink, the sum of the geometric multiplicities must be greater than or equal to $n$, that is:

By theoremlink, the sum of the algebraic multiplicities is equal to $n$, that is:

Substituting \eqref{eq:x90dgy3rmNu24WhyvLP} into \eqref{eq:TQhsAouxJRmwjztMdpm} gives:

Now, from theoremlink, we have that the algebraic multiplicity of an eigenvalue is greater than or equal to the geometric multiplicity of the same eigenvalue, that is:

Summing all of these inequalities gives:

Combining \eqref{eq:oo3zHU2Uji1nxu4Z8Dl} and \eqref{eq:SoPWDF64YY5gRAlrCBa} gives:

This means that:

Combining with \eqref{eq:x90dgy3rmNu24WhyvLP} gives:

This proves $(1)\implies(2)$.

We now prove $(2)\implies(3)$. Assume the sum of the geometric multiplicities is $n$. By theoremlink, the sum of the algebraic multiplicities is $n$, which means:

Again, by theoremlink, we have that:

Combining \eqref{eq:vCICCrzPdLZnYtErI9X} and \eqref{eq:DRURNqiWpvCY2V0roHb} implies:

This means that for every eigenvalue $\lambda_i$ for $i=1,2,\cdots,t$, the corresponding geometric multiplicity and algebraic multiplicity are equal. This proves $(2)\implies(3)$.

We now prove $(3)\implies(1)$. Suppose $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_t$ are distinct eigenvalues of an $n\times{n}$ matrix $\boldsymbol{A}$ where $t\le{n}$. Denote the corresponding eigenspace as $\mathcal{E}_1$, $\mathcal{E}_2$, $\cdots$, $\mathcal{E}_t$ and the corresponding basis as $\mathcal{B}_1$, $\mathcal{B}_2$, $\cdots$, $\mathcal{B}_t$ respectively.

We assume $(3)$ that the geometric multiplicity and algebraic multiplicity for every eigenvalue are equal. By theoremlink, because the sum of the algebraic multiplicities is equal to $n$, the sum of geometric multiplicities is also equal to $n$.

By theoremlink, because the basis $\mathcal{B}_1$, $\mathcal{B}_2$, $\cdots$, $\mathcal{B}_t$ correspond to distinct eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_t$, the union of the basis $\mathcal{B}_1\cup \mathcal{B}_2\cup \cdots\cup \mathcal{B}_t$ is linearly independent. This union can be expressed as:

Where $\boldsymbol{w}_i$ for $i=1,2,\cdots,n$ is a basis vector. The reason why the union contains $n$ vectors is that the sum of the geometric multiplicities is equal to $n$, which means that the total number of basis vectors when combined into a single set must be $n$.

We've now established that $\boldsymbol{A}$ has $n$ linearly independent eigenvectors. By theoremlink, we know that an $n\times{n}$ matrix with $n$ linearly independent eigenvectors is diagonalizable. Therefore, $\boldsymbol{A}$ is diagonalizable.

This completes the proof.

Solution. We have shown in examplelink that:

By theorem, we can refer to either of the following fact to prove that $\boldsymbol{A}$ is diagonalizable:

• the geometric multiplicity of the eigenvalue is n.

• the algebraic multiplicity and the geometric multiplicity are equal.