Worksheet 6

Solution.

1. To compute $T_{ij}$, we can just figure out what it does to the standard basis $\set{e_1, \ldots, e_n}$ of $\F^n$. If $k \notin \set{i, j}$, then the $i$-th and $j$-th rows of $e_k$ (or in other words, the $i$-th and $j$-th entries of $e_k$) are both $0$, so swapping those rows does nothing. I.e., $T_{ij} e_k = e_k$. However, if $k = i$, then we swap a $1$ in the $i$-th entry with a $0$ in the $j$-th entry, so $T_{ij} e_i = e_j$. Similarly, $T_{ij} e_j = e_i$. Thus, $T_{ij}$ looks like

   $$\begin{aligned} T_{ij} &= \begin{pmatrix} T_{ij}e_1 & \cdots & T_{ij}e_i & \cdots & T_{ij}e_j & \cdots & T_{ij}e_n \end{pmatrix} \\ &= \begin{pmatrix} 1 \\ & \ddots \\ & & 0 & \cdots & 1 \\ & & \vdots & \ddots & \vdots \\ & & 1 & \cdots & 0 \\ & & & & & \ddots \\ & & & & & & 1 \end{pmatrix}. \end{aligned}$$

   Most entries are $0$ except on the diagonal and the $1$'s the were moved off the diagonal. The off-diagonal $1$'s have indices $\p{i, j}$ and $\p{j, i}$.

   To compute the determinant, we perform Laplace expansion on the $i$-th column first. Assuming $i < j$, this expansion gives

   $$\det\p{T_{ij}} = \p{-1}^{j+i} \det \begin{pmatrix} 1 \\ & \ddots \\ & & 1 & \cdots & 1 \\ & & & \ddots & \vdots \\ & & & & 0 \\ & & & & & \ddots \\ & & & & & & 1 \end{pmatrix}.$$

   The matrix in the determinant is the minor we get from removing the $j$-th row and $i$-th column. Since the $j$-th row is below the top-right $1$, its row index doesn't change. However, because the $i$-column is to the left, we shifted it to the left by $1$. Thus, the $1$ has index $\p{i, j-1}$. Laplace expanding in the $\p{j-1}$-th column gives

   $$\det\p{T_{ij}} = \p{-1}^{j+i} \p{-1}^{i+\p{j-1}} \det \begin{pmatrix} 1 \\ & 1 \\ & & \ddots \\ & & & 1\end{pmatrix} = \p{-1}^{2j+2i-1} = -1.$$

   Here, we used the fact that $\det I_n = 1$.

   Next, $\det\p{T_{ij}^t} = \det\p{T_{ij}}$ since this is true for any matrix (or alternatively, you can say that $T_{ij}$ is symmetric, so it's equal to its transpose).

   Lastly, to show that $T_{ij}$ is its own inverse, we can just check it on a basis. If $k \notin \set{i, j}$, then

   $$T_{ij}\p{T_{ij}e_k} = T_{ij}e_k = e_k.$$

   And for $k = i$,

   $$T_{ij}\p{T_{ij}e_i} = T_{ij}\p{e_j} = e_i$$

   and similarly for $k = j$. Thus, $T_{ij}^2 = I_n$ holds on a basis, and by linearity, it holds everywhere, so $T_{ij}^{-1} = T_{ij}$.

2. Like before, we can compute $D_i\p{m}$ on a basis. If $k \neq i$, then $D_i\p{m}e_k = e_k$. Otherwise, $D_i\p{m}e_i = me_i$, so

   $$D_i\p{m} = \begin{pmatrix} 1 \\ & \ddots \\ & & m \\ & & & \ddots \\ & & & & 1 \end{pmatrix}.$$

   Here, $D_i\p{m}$ is $0$ except on the diagonal. At the $\p{i, i}$-th entry, it has an $m$, and it has a $1$ everywhere else on the diagonal. Since the determinant of an upper-triangular matrix is just the product of the diagonal entries (you can prove this by induction and Laplace expansion),

   $$\det\p{D_i\p{m}} = 1 \times \cdots \times m \times \cdots \times 1 = m.$$

   Like before, $\det\p{D_i\p{m}^t} = \det\p{D_i\p{m}}$ since it's symmetric. Lastly, if $m \neq 0$, then we can just check that $D_i\p{m^{-1}}$ is an inverse on a basis. If $k \neq i$, then

   $$D_i\p{m^{-1}}\p{D_i\p{m}e_k} = D_i\p{m^{-1}}e_k = e_k$$

   and if $k = i$,

   $$\begin{aligned} D_i\p{m^{-1}}\p{D_i\p{m}e_i} &= D_i\p{m^{-1}}\p{me_k} \\ &= m \p{D_i\p{m}^{-1}e_i} \\ &= mm^{-1}e_i \\ &= e_i. \end{aligned}$$

   Thus, $D_i\p{m^{-1}}$ is a left-inverse for $D_i\p{m}$, and because $D_i\p{m}$ is a square matrix, it follows that it's also a right-inverse. In other words, $D_i\p{m}^{-1} = D_i\p{m^{-1}}$.

3. I will assume that $i \neq j$. Otherwise, $L_{ii}\p{m} = D_i\p{1 + m}$, which is the previous case.

   When $k \neq j$, the $j$-th row of $e_k$ is $0$, so adding the $j$-th row of $e_k$ to anything doesn't change it. Thus, $L_{ij}\p{m}e_k = e_k$. On the other hand, if $k = j$, then we add $m$ to the $i$-th row of $e_k$, which gives $L_{ij}\p{m}e_j = e_j + me_i$. Hence,

   $$L_{ij}\p{m} = \begin{pmatrix} 1 \\ & \ddots \\ & & 1 & \cdots & m \\ & & & \ddots & \vdots \\ & & & & 1 \\ & & & & & \ddots \\ & & & & & & 1 \end{pmatrix}.$$

   Like before, every entry is $0$ except on the diagonal, which are all $1$'s, and except at the $\p{i, j}$-th entry, which is $m$. This is an upper-triangular matrix with all $1$'s on the diagonal, so its determinant is

   $$\det\p{L_{ij}\p{m}} = 1.$$

   Note that $L_{ij}\p{m}^t$ is a lower-triangular matrix, but the determinant of this matrix has the same property: it's the product of its diagonal entries, so $\det\p{L_{ij}\p{m}^t} = 1$ also.

   Lastly, if $k \neq j$, then

   $$L_{ij}\p{-m}\p{L_{ij}\p{m}e_k} = L_{ij}\p{-m}e_k = e_k,$$

   and if $k = j$,

   $$\begin{aligned} L_{ij}\p{-m}\p{L_{ij}\p{m}e_j} &= L_{ij}\p{-m}\p{e_j + me_i} \\ &= L_{ij}\p{-m}e_j + mL_{ij}\p{-m}e_i \\ &= e_j - me_i + me_i \\ &= e_j. \end{aligned}$$

   Thus, $L_{ij}\p{-m}$ is a left-inverse and $L_{ij}\p{m}$ is a square matrix, it is invertible with $L_{ij}\p{m}^{-1} = L_{ij}\p{-m}$.

Solution.

The definition of the determinant we're using is

where $i \in \set{1, \ldots, n}$ is any index, $\widehat{M}_{ij}$ is the minor of $M$ we get from removing the $i$-th row and $j$-th column (i.e., the row and column that contains $M_{ij}$). Thus, if $M$ is an $n \times n$ matrix, then $\widehat{M}_{ij}$ is an $\p{n-1} \times \p{n-1}$ matrix.

1. We prove this by induction. The base case is $n = 1$, where $M = \p{m_{11}}$. Then

   $$\det\p{\conj{M}} = \det\p{\conj{m_{11}}} = \conj{m_{11}} = \conj{\det\p{M}},$$

   so the base case holds. For the inductive step, assume that $\det\p{\conj{N}} = \conj{\det\p{N}}$ for any $n \times n$ matrix $N$. We need to show it holds for any $\p{n+1} \times \p{n+1}$ matrix. Let $M$ be an $\p{n+1} \times \p{n+1}$ matrix. Then

   $$\det\p{\conj{M}} = \sum_{j=1}^n \p{-1}^{i+j} \conj{M_{ij}} \det\p{\widehat{\conj{M_{ij}}}}.$$

   Note that $\widehat{\conj{M_{ij}}}$ is an $n \times n$ matrix, so by the inductive hypothesis,

   $$\det\p{\widehat{\conj{M_{ij}}}} = \conj{\det\p{\widehat{M_{ij}}}}.$$

   Also, recall that complex conjugation is additive and multiplicative: for $z, w \in \C$, we have $\conj{z + w} = \conj{z} = \conj{w}$ and $\conj{zw} = \conj{z} \conj{w}$. Lastly, note that $\conj{\p{-1}^{i+j}} = \p{-1}^{i+j}$ since it's a real number. Putting everything together,

   $$\begin{aligned} \det\p{\conj{M}} &= \sum_{j=1}^n \p{-1}^{i+j} \conj{M_{ij}} \det\p{\conj{\widehat{M}_{ij}}} \\ &= \sum_{j=1}^n \conj{\p{-1}^{i+j}} \conj{M_{ij}} \conj{\det\p{\widehat{M_{ij}}}} \\ &= \conj{\sum_{j=1}^n \p{-1}^{i+j} M_{ij} \det\p{\widehat{M_{ij}}}} \\ &= \conj{\det\p{M}}. \end{aligned}$$

2. Recall that $\p{A^t}_{ij} = A_{ji}$ for any matrix $A$. Thus,

   $$\p{\conj{M^t}}_{ij} = \conj{\p{M^t}_{ij}} = \conj{M_{ji}} = \p{\conj{M}}_{ji} = \p{\p{\conj{M}}^t}_{ij}.$$

3. Recall that if $A, B$ are square, then $\det\p{AB} = \det\p{A} \det\p{B}$. Thus, if we compute the determinant on both sides of $QQ^* = I_n$, then

   $$\det\p{Q} \det\p{Q^*} = \det\p{QQ^*} = \det\p{I_n} = 1.$$

   But by the first part,

   $$\det\p{Q^*} = \det\p{\conj{Q^t}} = \conj{\det\p{Q^t}} = \conj{\det\p{Q}}.$$

   Lastly, recall that if $z \in \C$, then $z\conj{z} = \abs{z}^2$, so

   $$1 = \det\p{Q} \det\p{Q^*} = \det\p{Q} \conj{\det\p{Q}} = \abs{\det\p{Q}}^2,$$

   which implies that $\abs{\det\p{Q}} = 1$.