Diagonal Matrices of Linear Operators

# Diagonal Matrices of Linear Operators

So far we have looked at linear operators $T \in \mathcal L (V)$ of finite-dimensional vector spaces $V$ for which certain bases $B_V$ of $V$ may yield an upper triangular matrix $\mathcal M (T, B_V)$. We will now look more into even simpler matrices for linear operators - diagonal matrices (which are themselves of course, upper triangular matrices). First let's look at the following definition.

Definition: If $V$ is a finite-dimensional vector space, then $T \in \mathcal L (V)$ is said to be Diagonalizable if there exists a basis $B_V$ such that $\mathcal M (T, B_V)$ is a diagonal matrix. |

The following proposition will tell us that if $T$ has $\mathrm{dim} (V)$ *distinct* eigenvalues, then there exists a basis $B_V$ of $V$ for which $\mathcal M (T, B_V)$ is a diagonal matrix.

Proposition 1: If $T \in \mathcal L (V)$ has $\mathrm{dim} (V) = n$ distinct eigenvalues then there exists a basis $B_V$ of $V$ for which $\mathcal M (T, B_V)$ is a diagonal matrix. |

**Proof:**Let $T \in \mathcal L(V)$ and suppose that $T$ has $\mathrm{dim} (V) = n$ distinct eigenvalues, call them $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbb{F}$. For each of these eigenvalues, choose a nonzero eigenvalue $v_j \in V$ that corresponds to $\lambda_j$ for $j = 1, 2, ..., n$. Now we know that these nonzero eigenvectors $v_1$, $v_2$, …, $v_n$ are linearly independent and there are $n$ of them, so $B_V = \{ v_1, v_2, ..., v_n \}$ forms a basis of $V$. Note that $T(v_j) = \lambda_jv_j$ for $j = 1, 2, ..., n$ and thus:

\begin{align} \quad \mathcal M(T, B_V) = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \end{align}

- Therefore $\mathcal M (T, B_V)$ is a diagonal matrix. $\blacksquare$

Now the following theorem will give us a bunch of equivalent statements if $T$ has a diagonal matrix with respect to some basis $B_V$ of $V$.

Theorem 1: If $T \in \mathcal L (V)$ for the finite-dimensional vector space $V$ and $\lambda_1, \lambda_2, ..., \lambda_m$ are distinct eigenvalues of $T$, then the following statements are equivalent:a) There exists a basis $B_V$ of $V$ such that $\mathcal M (T, B_V)$ is a diagonal matrix.b) $V$ has a basis consisting of corresponding nonzero eigenvectors from the eigenvalues $\lambda_1, \lambda_2, ..., \lambda_m$.c) There exists one-dimensional subspaces $U_1, U_2, ..., U_n$ of $V$ that are invariant under $T$ and such that $V = \bigoplus_{i=1}^{n} U_i$.d) $V = \bigoplus_{i=1}^{m} \mathrm{null} (T - \lambda_iI)$.e) $\mathrm{dim} V = \sum_{i=1}^{m} \mathrm{dim} ( \mathrm{null} (T - \lambda_iI))$. |

**Proof $a) \implies b)$:**This was proven in Proposition 1 above.

**Proof $b) \implies c)$:**Suppose that $V$ has a basis consisting of corresponding nonzero eigenvectors from the eigenvalues $\lambda_1, \lambda_2, ..., \lambda_m$, and let $\{ v_1, v_2, ..., v_n \}$ be this basis of eigenvectors. Let $U_j = \mathrm{span} (v_j)$ for $j = 1, 2, ..., n$. Then clearly $V = \sum_{i=1}^{n} U_i$.

- Each $U_j$ is a one-dimensional subspace of $V$. To show that each $U_j$ is invariant under $T$, let $u \in U_j = \mathrm{span} (v_j)$. Then for some $a \in \mathbb{F}$ we have that $u = av_j$. So $T(u) = T(av_j) = aT(v_j) = a \lambda v_j \in \mathrm{span} (v_j) = U_j$.

- Now since $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$, we have that this basis is linearly independent and so for every $v \in V$ we have that for some scalars $a_1, a_2, ..., a_n \in \mathbb{F}$ that $v$ can be written uniquely as a linear combination of these basis vectors:

\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}

- Since $U_j = \mathrm{span} (v_j)$ for $j = 1, 2, ..., n$ we have that $v$ can be uniquely written as:

\begin{align} \quad v = \underbrace{u_1}_{\in U_1} + \underbrace{u_2}_{\in U_2} + ... + \underbrace{u_n}_{\in U_n} \end{align}

- Therefore $V = \bigoplus_{i=1}^{n} U_i$.

**Proof of $c) \implies b)$:**Suppose that there exists one-dimensional subspaces $U_1, U_2, ..., U_n$ such that each is invariant under $T$ and $V = \bigoplus_{i=1}^{n} U_i$. Let $v_j \in U_j$ be a nonzero vector for each $j = 1, 2, ..., n$. Each of these vectors $v_j$ is an eigenvector of $T$. Now since $V = \bigoplus_{i=1}^{n} U_i$, we have that $V = \sum_{i=1}^{n} U_i$ and so every $v \in V$ can be written uniquely as a sum $v = u_1 + u_2 + ... + u_n$ where $u_j \in U_j$ for $j = 1, 2, ..., n$. But $u_j$ is a scalar multiple of the nonzero vectors $v_j$ since $v_j \in U_j$ and $U_j$ is one-dimensional, and so $\{ v_1, v_2, ..., v_n \}$ must be a basis of $V$.

**Proof of $b) \implies d)$:**Suppose that $V$ has a basis of corresponding nonzero eigenvectors from the eigenvalues $\lambda_1, \lambda_2, ..., \lambda_m$, call it $\{v_1, v_2, ..., v_n \}$. Then if $v \in V$ we have that $v$ is a linear combination of these basis vectors. But the set of linear combinations of each of these eigenvalues is $\mathrm{null} (T - \lambda_jI)$ for $j = 1, 2, ..., n$ and so

\begin{align} V = \sum_{i=1}^{m} \mathrm{null} (T - \lambda_iI) \end{align}

- We now need to show that this sum is direct. Suppose that $0 = u_1 + u_2 + ... + u_m$ where $u_j \in \mathrm{null} (T - \lambda_jI)$ for $j = 1, 2, ..., m$. Now since $u_j \in \mathrm{null} (T - \lambda_jI)$, then $u_j$ is an eigenvector of $\lambda_j$. But $\lambda_1, \lambda_2, ..., \lambda_m$ are distinct eigenvalues and so this implies that $u_1 = u_2 = ... = u_m = 0$, and so $V = \bigoplus_{i=1}^{m} \mathrm{null} (T - \lambda_iI)$.

**Proof of $d) \implies e)$:**Suppose that $V = \bigoplus_{i=1}^{m} \mathrm{null} (T - \lambda_i)$. Since $V$ is finite-dimensional and $\mathrm{null} (T - \lambda_i)$ for $i = 1, 2, .., m$ are subspace of $V$, it follows immediately that $\mathrm{dim} V = \mathrm{dim} ( \mathrm{null} (T - \lambda_1I)) + \mathrm{dim} ( \mathrm{null} (T - \lambda_2I)) + ... + \mathrm{dim} ( \mathrm{null} (T - \lambda_mI))$.

**Proof of $e) \implies b)$:**Suppose that $n = \mathrm{dim} (V) = \sum_{i=1}^{m} \mathrm{dim} ( \mathrm{null} (T - \lambda_iI))$. Since $\mathrm{null} (T - \lambda_iI)$ for $j = 1, 2, ..., m$ is a subspace of $V$, then choose a basis for each $\mathrm{null} (T - \lambda_jI)$, and take all of these bases to form a set of $n$ eigenvectors $\{ v_1, v_2, ..., v_n \}$ of $T$. We want to show that this set of vectors is linearly independent to show that it is a basis of $V$.

- Suppose that for $a_1, a_2, ..., a_n \in \mathbb{F}$ we have that $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$. Then for $j = 1, 2, ..., m$ let $u_j$ be equal to the sum of terms $a_kv_k$ such that $v_k \in \mathrm{null}(T - \lambda_jI)$. Each of these $u_j$ is an eigenvector of $T$ that corresponds to the distinct eigenvalue $\lambda_j$ and so $u_1 + u_2 + ... + u_m = 0$. But $u_1 = u_2 = ... = u_m = 0$ since $\lambda_1, \lambda_2, ..., \lambda_m$ are distinct eigenvalues.

- Now since $u_j$ is equal to the sum of terms $a_kv_k$ such that $v_k \in \mathrm{null} (T - \lambda_jI)$, then the corresponding bases of $v_k$'s implies that all of the terms $a_k$ are equal to zero. Thus $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of $n$ eigenvectors, so $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$. $\blacksquare$