# square, symmetric and skew-symmetric matrices, eigenspace

Could someone help me with question 5; maybe give me a hint on how to approach.

Let n be a fixed fixed integer and let $V$ be the vector space of $(n×n)$-matrices.
A matrix $\;A$ is symmetric if $\;A=A^T$, and skew-symmetric if $\;A=−A^T$

1. Show that the subset of symmetric matrices $W_1\subseteq V$, and skew-symmetric matrices $W_2 \subseteq V$ are subspaces.

2. Show that $W_1\cap W_2=0$, and that $W_1+W_2=V$.

3. Show that the transformation $\;T:V\to V$, where $T(A)=A+A^T$, is linear.

4. Show that $W_1$ and $W_2$ are eigenspaces of $T$.

5. Show that there exists a basis of $V$, such that the matrix representation of the transformation $T$ in that basis, is a diagonal matrix.

1. A subset $\;H⊆V$ is a subspace if:
• $\mathbf{0}\in H$
• $\mathbf{u,v}∈H \Leftrightarrow (\mathbf{u}+\mathbf{v}) \in H)$
• $\mathbf{u}∈H \Leftrightarrow (c\mathbf{u})\in H, c\in R$.

Symmetric matrices $W_1\subseteq V$:
$\mathbf{0}^T=\mathbf{0} \Leftrightarrow \mathbf{0}\in W_1$.
Let $\;B_1, B_2\in V$ be symmetric matrices. Then by definition $\;B_1=B_1^T$ and $\;B_2=B_2^T$. $(B_1+B_2)^T=B_1^T+B_2^T=B_1+B_2=(B_1+B_2) \implies (B_1+B_2) \in W_1$.
Let $\;c \in R$. $(cB)^T=c(B^T)=cB \implies (cB)\in W_1$

Skew-symmetric matrices $W_2\subseteq V$:
$\mathbf{0}^T=\mathbf{0}=\mathbf{-0} \Leftrightarrow \mathbf{0}\in W_2$.
Let $\;C_1, C_2\in V$ be skew-symmetric matrices. Then by definition $\;C_1=-C_1^T$ and $\;C_2=-C_2^T$. $(C_1+C_2)^T=C_1^T+C_2^T=-C_1-C_2=-(C_1+C_2) \implies (C_1+C_2) \in W_2$.
Let $\;c \in R$. $(cC)^T=c(C^T)=c(-C)=-(cC) \implies (cC) \in W_2$

1. Let $A \in V$ be an arbitrary $(n×n)$-matrix. Then $A\in W_1 \Leftrightarrow A^T=A$ and $A\in W_2 \Leftrightarrow A^T=-A$. Therefore, $\;A\in (W_1\cap W_2) \implies A=-A \implies A=0$. $W_1\cap W_2=0$

Let $A\in V, B\in W_1 and C\in W_2$. Then $B^T=B$ and $C^T=-C$.
Suppose that $W_1+W_2=V$. Then $\;A=B+C$ which implies that $A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$.

We now have that $\begin{cases}A=B+C \\ A^T=B-C\end{cases} \implies \begin{cases}B=\frac{A+A^T}{2} \\ C=\frac{A-A^T}{2}\end{cases} \implies A = B+C=\frac{A+A^T}{2}+\frac{A-A^T}{2} \Leftrightarrow A=\frac{A+A^T+A-A^T}{2}=\frac{2A}{2}=A$. This proves that any arbitrary $(n×n)$-matrix can be written as the sum of a symmetric matrix and a skew-symmetric matrix, i.e. $W_1+W_2=V$

The transformation $\;T:V\to V$, where $T(A)=A+A^T$, is linear if and only if:
- $\;T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v}), \forall \mathbf{u}, \mathbf{v} \in V$
- $\;T(c\mathbf{u})=cT(\mathbf{u})), \forall \mathbf{u}\in V, \forall c\in R$
In (2), we showed that any arbitrary $(n×n)$-matrix can be written as the sum of a symmetric matrix and a skew-symmetric matrix, i.e. $A=B+C, \forall A \in V, \forall B \in W_1, \forall C \in W_2$.

$T(A_1+A_2)=T(A_1)+T(A_2)=T(B_1+C_1)+T(B_2+C_2)=T(B_1)+T(C_1)+T(B_2)+T(C_2)=B_1+B_1^T+C_1+C_1^T+B_2+B_2^T+C_2+C_2^T=B_1+B_1+C_1-C_1+B_2+B_2+C_2-C_2=2B_1+2B_2=2(B_1+B_2)$ In (2), we showed that any arbitrary symmetric matrix $B$ can be written as $B=\frac{A+A^T}{2}$. So, $T(A_1+A_2)=2(B_1+B_2)=2(\frac{A_1+A_1^T}{2}+\frac{A_2+A_2^T}{2})=2\frac{A_1+A_1^T+A_2+A_2^T}{2}=(A_1+A_2)+(A_1^T+A_2^T)=(A_1+A_2)+(A_1+A_2)^T$

$T(cA)=cA+(cA)^T=cA+cA^T=c(A+A^T)$

1. If a transformation $T$ has vectors $\mathbf{x}$, for which $T(\mathbf{x})=\lambda \mathbf{x}$, for some skalar $\lambda$, then these vectors $\mathbf{x}$ are called eigenvectors corresponding to that eigenvalue $\lambda$. Furthermore, if a subspace $H$ is the span of these eigenvectors, then $H$ is an eigenspace of $T$ for that eigenvalue.

Let $B\in W_1$ and $C\in W_2$.
$T(B)=B+B^T=B+B=2B \implies \lambda=2$
$T(C)=C+C^T=C-C=0C \implies \lambda=0 \Leftrightarrow W_2 \subseteq Nul(A)$

Could someone help me with question 5; maybe give me a hint on how to approach.

Hint: take an element $S$ of $W_1$; apply $T$ to it. How is $T(S)$ related to $S$?
Do the same for $W_2$.
• You've shown that every vector in $W_1$ is an eigenvector for $\lambda = 2$, and every verctor in $W_2$ is an eigenvector for $\lambda= 0$. So both are eigenspaces. Suppose you picked a basis for $W_1$ and a basis for $W_2$...would they together make a basis for all of $V$? – John Hughes Nov 25 '17 at 20:05
• I did give you a hint: I asked you "Suppose you picked a basis for $W_1$ and a basis for $W_2$...would they together make a basis for all of $V$?" When you've answered that, let's talk. – John Hughes Nov 26 '17 at 13:12
• Yes, if I picked a basis for $W_1$ and a basis for $W_2$, they would together make a basis for all of $V$, since a basis for a subspace is a set of linearly independent vectors that span the subspace. If $B={b_1, b_2, ..., b_n}$ is a basis for $W_1$ and $C={c_1, c_2, ..., c_n}$ is a basis for $W_2$, V=Span{b_1, b_2, ..., b_n, c_1, c_2, ..., c_n}. Right? – Filip Nov 26 '17 at 20:05