Problem # A.2:
An \(m\)-by-\(n\) matrix with coefficients in a field \(\mathbb{F}\)
is defined to be an \(m\)-by-\(n\) array of elements of \(\mathbb{F}\). We
write \(M_{m,n}(\mathbb{F})\) for the set of all such matrices, so an
element \(A\in M_{m,n}(\mathbb{F})\) looks like
\[
A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end{pmatrix}
\]
We make \(M_{m,n}(\mathbb{F})\) into a vector space in
the obvious way:
\[
\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn} \\
\end{pmatrix}
+
\begin{pmatrix}
b_{11} & \cdots & b_{1n} \\
\vdots & \ddots & \vdots \\
b_{m1} & \cdots & b_{mn} \\
\end{pmatrix}
=
\begin{pmatrix}
a_{11}+b_{11} & \cdots & a_{1n}+b_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1}+b_{m1} & \cdots & a_{mn}+b_{mn} \\
\end{pmatrix}
\]
and
\[
c \begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn} \\
\end{pmatrix}
=
\begin{pmatrix}
ca_{11} & \cdots & ca_{1n} \\
\vdots & \ddots & \vdots \\
ca_{m1} & \cdots & ca_{mn} \\
\end{pmatrix}
\]
Problem # A.3:
The transpose of a (square)
matrix \(A\), denoted \(A^*\), is obtained
by flipping the entries across the main diagonal. So for example
\[
\begin{pmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\ \end{pmatrix}^*
=
\begin{pmatrix} 1&4&7\\ 2&5&8\\ 3&6&9\\ \end{pmatrix}.
\]
A matrix \(A\) is symmetric if \(A^*=A\)
and it is anti-symmetric if \(A^*=-A\).
Return to the
Math 540 Homework Page
Return to the
Math 540 Home Page
(a)
Write down a basis for the \(\mathbb{F}\)-vector space
\(M_{3,2}(\mathbb{F})\) of 3-by-2 matrices. What is the dimension of
\(M_{3,2}(\mathbb{F})\)?
(b)
More generally, what is the dimension of
\(M_{m,n}(\mathbb{F})\)?
Solution:
(a) The following six vectors form a basis for \(M_{3,2}(\mathbb{F})\),
so \(M_{3,2}(\mathbb{F})\) has dimension 6.
\[
\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{pmatrix},\quad
\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{pmatrix},\quad
\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ \end{pmatrix},\quad
\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix},\quad
\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ \end{pmatrix},\quad
\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{pmatrix}.
\]
To see this, note that
\[
\begin{pmatrix} a & b \\ c & d \\ e & f \\ \end{pmatrix}=
a\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{pmatrix}+
b\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{pmatrix}+
c\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ \end{pmatrix}+
d\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix}+
e\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ \end{pmatrix}+
f\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{pmatrix}.
\]
This shows that every matrix in \(M_{3,2}(\mathbb{F})\) is a linear
combination of the given 6 matrices, so the given
6 matrices span. Further, it shows that the only
linear combination of the 6 matrices that equals the 0 matrix is by taking the
0 multiple of each of them, so the 6 matrices are linearly indepedent.
(b)
Let \(E_{ij}\) be the matrix that has a \(1\) for its entry in
the \(i\)'th row and \(j\)'th column, and every other entry is 0.
Then
\[
\{ E_{ij} : 1\le i\le m,\; 1\le j\le n \}
\]
is a basis for \(M_{m,n}(\mathbb{F})\), so \(M_{m,n}(\mathbb{F})\) has
dimension \(mn\). To see that this is a basis, we note that
\[
\sum_{i=1}^m \sum_{j=1}^n a_{ij}E_{ij} =
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end{pmatrix},
\]
so we get every matrix in \(M_{m,n}(\mathbb{F})\) as a linear
combination of the \(E_{ij}\) matrices, so the \(E_{ij}\) matrices span,
and the only
linear combination of the \(E_{ij}\) matrices
that equals the zero matrix is taking all of the
\(a_{ij}\) to be 0, so the \(E_{ij}\) matrices are linearly independent.
(a)
Prove that the set of \(n\)-by-\(n\) symmetric matrices is a
vector subspace of \(M_{n,n}(\mathbb{F})\).
(b)
Find a basis for the space of 2-by-2 symmetric matrices. What
is its dimension?
(c)
Generalize by describing a basis for the space of \(n\)-by-\(n\) symmetric
matrices and computing its dimension. (It might help to start with 3-by-3.)
(d)
Prove that the set of
\(n\)-by-\(n\) anti-symmetric matrices is also a vector subspace
of \(M_{n,n}(\mathbb{F})\),
describe a basis, and compute its dimension.
(e) (Bonus)
Let's write \(M_{n,n}(\mathbb{F})^{\text{sym}}\)
for the space of symmetric matrices and
\(M_{n,n}(\mathbb{F})^{\text{anti-sym}}\)
for the space of anti-symmetric matrices. Prove that
\[
M_{n,n}(\mathbb{F})
= M_{n,n}(\mathbb{F})^{\text{sym}} + M_{n,n}(\mathbb{F})^{\text{anti-sym}}.
\]
Is this a direct sum of vector spaces?
Solution:
What happens if we add two matrices and then take their transpose?
We compute
\[
\begin{align*}
(A+B)^*
&= \left(\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn} \\
\end{pmatrix}
+
\begin{pmatrix}
b_{11} & \cdots & b_{1n} \\
\vdots & \ddots & \vdots \\
b_{m1} & \cdots & b_{mn} \\
\end{pmatrix}\right)^* \\
&=
\begin{pmatrix}
a_{11}+b_{11} & \cdots & a_{1n}+b_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1}+b_{m1} & \cdots & a_{mn}+b_{mn} \\
\end{pmatrix}^* \\
&=
\begin{pmatrix}
a_{11}+b_{11} & \cdots & a_{m1}+b_{m1}\\
\vdots & \ddots & \vdots \\
a_{1n}+b_{1n} & \cdots & a_{mn}+b_{mn} \\
\end{pmatrix} \\
&= \left(\begin{pmatrix}
a_{11} & \cdots & a_{m1} \\
\vdots & \ddots & \vdots \\
a_{1n} & \cdots & a_{mn} \\
\end{pmatrix}
+
\begin{pmatrix}
b_{11} & \cdots & b_{m1} \\
\vdots & \ddots & \vdots \\
b_{1n} & \cdots & b_{mn} \\
\end{pmatrix}\right) \\
&= A^* + B^*
\end{align*}
\]
and similarly \((cA)^* = c A^*\). So we have the useful formulas
\[
(A+B)^*=A^*+B^*\quad\text{and}\quad(cA)^* = c A^*.
\]
(a)
Suppose that \(A\) and \(B\) are
symmetric matrices, so \(A^*=A\) and \(B^*=B\). Then
using the formulas that we just proved, we have
\[
(A+B)^* = A^*+B^* = A+B
\quad\text{and}\quad
(cA)^* = cA^* = cA.
\]
This shows that the set of symmetric
matrices \(M_{n,n}^{\text{sym}}(\mathbb{F})\) is closed under addition
and scalar multiplication, so it is a vector subspace
of \(M_{n,n}(\mathbb{F})\).
(b)
Consider the three matrices
\[
E_{11} = \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix},\quad
E_{22} = \begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix},\quad
F_{12} = \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix}.
\]
Then any 2-by-2 symmetric matrix can be written as
\[
\begin{pmatrix} a&b\\ b&d\\ \end{pmatrix}
= \begin{pmatrix} a&0\\ 0&0\\ \end{pmatrix}
+ \begin{pmatrix} 0&0\\ 0&d\\ \end{pmatrix}
+ \begin{pmatrix} 0&b\\ b&0\\ \end{pmatrix}
= aE_{11} + dE_{22} + bF_{12}
\]
in exactly one way, so \(\{E_{11},E_{22},F_{12}\}\) is a basis
for \(M_{n,n}^{\text{sym}}(\mathbb{F})\). In particular,
\(\dim M_{n,n}^{\text{sym}}(\mathbb{F})=3\).
(c)
More generally, as in Problem A.2, let \(E_{ij}\) be the matrix that
has a \(1\) for its entry in the \(i\)'th row and \(j\)'th column, and
every other entry is 0. Also let
\[
F_{ij} = E_{ij} + E_{ji},
\]
so \(F_{ij}\) has two 1's and the rest of its entries are 0. Since
it's clear that
\[
E_{ij}^* = E_{ji},
\quad\text{we have}\quad
F_{ij}^* = (E_{ij} + E_{ji})^*
= E_{ij}^* + E_{ji}^*
= E_{ji} + E_{ij}
= F_{ij},
\]
so \(F_{ij}\) is a symmetric matrix. It's also clear that
\(E_{ii}\) is a symmetric matrix. Then generalizing the proof
in (b), one sees that
\[
\{E_{ii} : 1\le i\le n\}
\cup
\{F_{ij} : 1\le i\lt j\le n\}
\]
is a basis for \(M_{n,n}^{\text{sym}}(\mathbb{F})\). (Note that
\(F_{ij}=F_{ji}\), so we only need one of them.)
More precisely, since a symmetric matrix has \(a_{ij}=a_{ji}\), we have
\[
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{pmatrix}
= \sum_{i=1}^n a_{ii}E_{ii}
+ \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_{ij}F_{ij}.
\]
Finally,
\[
\dim M_{n,n}^{\text{sym}}(\mathbb{F})
= n + \frac{n(n-1)}{2} = \frac{n^2+n}{2}.
\]
(d)
This is similar, but instead of using the \(E_{ij}\) and
the \(F_{ij}\), use the matrices \(G_{ij}=E_{ij}-E_{ji}\).
Note that
\[
G_{ij}^* = (E_{ij}-E_{ji})^*
= E_{ij}^*-E_{ji}^*
= E_{ji}-E_{ij}
= -G_{ij},
\]
so \(G_{ij}\in M_{n,n}^{\text{anti-sym}}(\mathbb{F})\).
Further, it's not hard to show
that \(\{G_{ij} : 1\le i\lt j\le n\}\) is a basis for
\(M_{n,n}^{\text{anti-sym}}(\mathbb{F})\).
Thus a matrix is anti-symmetric if and only if \(a_{ij}=-a_{ji}\),
which means that the matrix can be written as
\[
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{pmatrix}
= \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_{ij}G_{ij}.
\]
(Notice in particular that diagonal entries satisfy \(a_{ii}=0\),
since they equal their own negatives.) Hence
\[
\dim M_{n,n}^{\text{anti-sym}}(\mathbb{F})= \frac{n^2-n}{2}.
\]
(e)
Given any matrix \(A\in M_{n,n}(\mathbb{F})\), write \(A\) as
\[
A = \frac{1}{2}(A+A^*) + \frac{1}{2}(A-A^*).
\]
Then it's easy to check that \(\frac{1}{2}(A+A^*)\) is symmetric and
\(\frac{1}{2}(A-A^*)\) is anti-symmetric, which shows the desired sum
formula. In order to show that it is a direct sum, it's enough to
show (by a result in the book) that the only matrix that is both symmetric
and anti-symmetric is the zero matrix. But if \(A\) is both symmetric
and anti-symmetric, then
\[
A = A^* = -A,\quad\text{so}\quad 2A=0,\quad\text{so}\quad A=0.
\]