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:

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 linear systems with diagonal matrices \begin{equation} \mqty(5 & 0 \ 0 & 2) \mqty(c_1 \ c_2) = \mqty(10 \ 1) \end{equation} Each output is simply multiplying a value by th scaling factor corresponding to that row. 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: