The diagonal of a square matrix consists of entries from the upper-left to the bottom-right

Furthermore, because eigenvalues of a map are the entries of the diagonal of its upper-triangular matrix, and this is *technically* an upper-triangular matrix, the entries on the diagonal are exactly the eigenvalues of the Linear Map.

## properties of diagonal matrices

Suppose \(V\) is finite-dimensional, and \(T \in \mathcal{L}(V)\); and let \(\lambda_{1}, … \lambda_{m}\) be distinct eigenvalues of \(T\). Then, the following are equivalent:

- \(T\) is diagonalizable
- \(V\) has a basis containing of eigenvectors of \(T\)
- there exists 1-dim subspaces \(U_{1}, …, U_{n}\) of \(V\), each invariant under \(T\), such that \(V = U_1 \oplus … \oplus U_{n}\)
- specifically, those \(U\) are eigenspaces; that is: \(V = E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T)\)
- \(\dim V = \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T)\)

Proof:

\(1 \implies 2\), \(2 \implies 1\)

By a moment’s fucking thought. hehe my notes my rule. jkjkjk

By calculation this is true; if you apply a standard basis to the matrix, it will simply be scaled; therefore, you can think of each slot as an eigenvector of \(T\).

\(2 \implies 3\)

Create \(U_{j} = span(v_{j})\) where \(v_{j}\) is the \(j\) eigenvalue of \(T\). Now, given \(v_{j}\) forms a basis, then, \(v_1 … v_{n}\) not only is linearly independent but span. Therefore, each vector in \(V\) can be written uniquely by a linear combination of \(v_{j}\) (i.e. taking one \(v_{j}\) from each \(U\)). Hence, by definition, \(U_{j}\) form a direct sum to \(V\), hence showing \(3\).

\(3\implies 2\)

Now, suppose you have a bunch of 1-d invariant subspaces \(U_1 … U_{n}\) and they form a direct sum; because they are invariant subspaces, picking any \(v_{j} \in U_{j}\) would be an eigenvector (because \(T v_{j} = a_{j} v_{j}\), as applying \(T\) is invariant so it’d return to the same space, just at a different place). Now, because they form a direct sum on \(V\), taking \(v_{j}\) from each \(U_{j}\) would result in a linearly independent list which—because they sum up to $V$—span all of \(V\) as each \(U\) is simply spanning by scaling \(v_{j}\). So, \(v_{j}\) together forms a basis.

\(2 \implies 4\)

Given \(V\) has a basis formed by eigenvectors of \(T\), the sum of all scales of eigenvectors in \(T\) can be written by the sum of all eigenspaces: that is \(V = null(T-\lambda_{1} I) + … null(T- \lambda_{m} I)\) (recall that \(E(\lambda_{j}, T) = null(T- \lambda_{j}I)\)); as each eigenvalue for which the basis is formed can be found in each of these spaces, their sum would therefore equal to \(V\) as this sum represents an linear combination of eigenvectors in \(T\).

Now, sum of eigenspaces form a direct sum so we have that the sum is direct sum. Hence: \(V = E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T)\)

\(4 \implies 5\)

\(U_1 + \dots + U_{m}\) is a direct sum IFF \(\dim (U_1 + \dots + U_{m}) = \dim U_1 + \dots + \dim U_{m}\) (see link for proof).

\(5 \implies 2\)

We are given that:

\begin{equation} \dim V = \dim E(\lambda_{1}, T) + \dots + \dim E(\lambda_{m}, T) \end{equation}

which means that taking a basis of each subspace provides a list of \(\dim n\) long of eigenvectors. Now, each sub-list belonging to each space is linearly independent amongst themselves, and they will each be linearly independent against others as list of eigenvectors are linearly independent.

i.e.: if \(a_1v_1 + … + a_{n} v_{n} = 0\), we can treat each chunk from each eigenspace as \(u\), making \(u_1 + … u_{m} = 0\); as they are eigenvectors from distinct eigenvalues, they are linearly independent so each will be \(0\). Now, collapsing it into the basis of each eigenspace, this makes \(a_{j}\) of the coefficients \(0\) as well.

And all of this makes \(v_1 … v_{n}\) a list of \(\dim n\) long that is linearly independent; hence, it is a basis of \(V\), as desired. \(\blacksquare\)

## enough eigenvalues implies diagonalizability

If \(T \in \mathcal{L}(V)\) has \(\dim V\) distinct eigenvalues, then \(T\) is diagonalizable.

Proof:

let \(\dim V = n\). Pick eigenvectors \(v_1 … v_{n}\) corresponding to distinct eigenvalues \(\lambda_{1} … \lambda_{n}\). Now, eigenvectors coorsponding to distinct eigenvalues are linearly independent, and this is a list of \(\dim n\) long that is linearly independent, so it is a basis of eigenvectors. Now, that means that the matrix coorsponding to \(T\) is diagonalizable.

**NOTE THAT THE CONVERSE IS NOT TRUE!** as each eigenspace can have a dimension of more than 1 so 1 eigenvalue can generate two linearly independent eigenvectors belonging to it.

For instance:

\begin{equation} T (z_1, z_2, z_3) = (4z_1, 4z_2, 5z_3) \end{equation}