# Algebra (Artin) Ch.6 Problem 6.10

Problem statement: Let $M$ be a matrix made up of two diagonal blocks: $M = \begin{bmatrix} A & 0 \\ 0 & D \end{bmatrix}$. Then $M$ is diagonalizable if and only if $A$ and $D$ are diagonalizable.

Solution: Let $A, D$ be $n \times n$ matrices. Then $M$ is an $2n \times 2n$ matrix. Let $T$ be the linear operator and $(u_1,..,u_n,u'_1,..,u'_n)$ be the basis of the vector space $F^{2n}=V$ w.r.t the matrix $M$. Denote $Sp({u_1,..,u_n})$ by W and $Sp({u'_1,..,u'_n})$ by W’. Then W and W’ are T-invariant subspaces. Also, $V$ is the direct sum of $W$ and $W'$.

The if part:

Let $M$ be diagonalizable. Then there is a basis of eigenvectors of the vector space $F^{2n}$. Let the basis be $(v_1,..,v_{2n}).$ Then $v_i = w_i + w'_i$ for some $w_i \in W$ and $w'_i \in W'$. Then $T(v_i)=\lambda_i v_i = \lambda_i w_i + \lambda_i w'_i = T(w_i) + T(w'_i)$.

Then $T(w_i)-\lambda_i w_i + T(w'_i)-\lambda_i w'_i = 0$. Since W and W’ are T-invariant, $T(w_i)=\lambda_i w_i, T(w'_i)=\lambda_i w'_i$. Then the nonzero $w_i s$ and $w'_i s$ are eigenvectors of W and W’ respectively. Also, since V is the direct sum of W and W’, and $(v_1,..,v_{2n})$ is a basis of V, $(w_1,..,w_n)$ generates W and $(w'_1,..,w'_n)$ generates W’. Then we can obtain a basis of eigenvectors for W and W’ by dropping the dependent vectors in $(w_1,..,w_n)$ and $(w'_1,..,w'_n)$. This essentially means that A and D are diagonalizable.

The only-if part:

As $A, D$ are diagonalizable, let $P^{-1}AP$ and $Q^{-1}DQ$ be diagonal matrices where $P,Q$ are invertible. Then $\begin{bmatrix} P & 0 \\ 0 & I_n \end{bmatrix}$ $\begin{bmatrix} I_n & 0 \\ 0 & Q \end{bmatrix}$ diagonalizes $M$.

# Algebra (Artin) Ch.4 Exercise 6.8

Problem statement: A linear operator is nilpotent if some positive power $T^k$ is zero. Prove that $T$ is nilpotent if and only if there is a basis of $V$ such that the matrix of $T$ is upper triangular, with diagonal entries zero.

Solution: (assume that V is finite-dimensional) Let the matrix of $T$ be $A$.

The if part:

Let $k$ be the smallest positive integer such that $T^k$ = 0. Then $ker T \neq {0},$ otherwise $dim V=dim (Im T)$, which would mean $A$ is invertible, a contradiction.

Let us denote $ker T^i$ by $K_i$, to save time.

If $T^i(v)=0$, then $T^{i+1}(v)=0$, so $K_i \subset K_{i+1}$. We note that $K_k=V$.

Then, a basis for $K_1$ can be extended to a basis for $K_2$, which can be extended to a basis for $K_3$ and so on, eventually we shall obtain a basis for $K_k$, or $V$. Since $k$ is the smallest integer such that $T^k=0,$ we have $K_i \neq K_j$ whenever $i \neq j, 0 Let this basis be $(u_11,..,u_1r_1,u_21,..,u_2r_2,...,u_k1,..,u_kr_k),$ where $(u_11,..,u_ir_i)$ is a basis for $K_i$.

Let $M$ be the matrix of $T$ with respect to this basis. Then the first $r_1$ columns of $M$ are zero, the next $r_2$ columns have the last $n-r_1$ entries zero, the next $r_3$ columns have the last $n - (r_1+r_2)$ entries zero and so on. This is because $T(K_{i+1}) \subset K_i$. Clearly, $M$ is an upper triangular matrix. We would get a lower triangular matrix if we wrote the basis vectors in the opposite order.

Only if part:

Let $A$ be upper triangular and $(v_1,..,v_n)$ be the basis w.r.t which the matrix of $T$ is $A$. Then $T(v_1)=0, T(v_2)=a_12 v_1,$ so $T^2(v_2)=0,$ proceeding this way, we see that $T^i(v_i)=0, i=1,2,..,n$, so $T^n=0.$

# Michael Artin’s Algebra Ch.3 M.3(c)

Problem Statement: prove that every pair $x(t), y(t)$ of real polynomials satisfies some real polynomial relation $f(x,y)=0$.

Solution sketch:

Let, without loss of generality, $x=a_0 + a_{1} t + a_{2} t^2+...+a_{n} t^n, a_{n} \neq 0,$ $y = b_0 + b_{1} t +...+ b_{m} t^m, b_{m} \neq 0, m \le n.$

Then $t^{m}$ can be written as a linear combination of  $y, t, t^{2},.., t^{m-1}$. So, in $x(t)$, we write $t^{k}$ as $t^{m}t^{k-m}$ when $m \le k$, and replace $t^{m}$ by $y - (b_0 + b_{1} t +...+ b_{m-1} t^{m-1}$. By repeating this process as long as there are powers of $t$ greater than or equal to $m$ in $x(t)$, we shall eventually find an equation where the highest power of $t$ is no greater than $m-1$. Now let the highest power of $t$ in this equation be $r$. Then in a similar fashion, we can plug in the value of $t^{r}$ in $y(t)$ to obtain an equation where the power of $t$ will not exceed $r - 1$. By repeating this process, we shall eventually obtain an equation where the power of $t$ is zero, and that is our desired $f(x,y)=0$.

# Michael Artin’s Algebra Ch.1 M.1

(I have started reading Michael Artin’s Algebra and I think uploading solutions of some problems, especially some of the starred ones, might be useful to others like me who are currently self-studying mathematics. If I have obtained the solution from some source, I will mention the source with the solution. Pointing out errors is always very welcome.)

Problem Statement: $M= \begin{bmatrix} A & B \\ C & D \end{bmatrix}$ where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$ . Use block multiplication to prove that $det M=det(AD-CB)$.

Solution:

We note that $G=\begin{bmatrix} 0 & I_n \\ A & -C \end{bmatrix}$ is invertible, and by expanding the determinant along its first row, we see that $det G = det A$.

Then $GM= \begin{bmatrix} A & B \\ 0 & AD-CB \end{bmatrix}$.

Take $H=\begin{bmatrix} A^{-1} & 0 \\ 0 & I_n \end{bmatrix}$.

Then $HGM = \begin{bmatrix} I_n & A^{-1}B \\ 0 & AD-CB \end{bmatrix}$.

Clearly, $det M = det (HGM)$. Expanding $HGM$ along its first column, we get $det (HGM) = det (AD - CB)$.