1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Hint: take an element $S$ of $W_1$; apply $T$ to it. How is $T(S)$ related to $S$?

Do the same for $W_2$.

(Go ahead and put your answers to these in as partial results for step 4 in your question, using the "edit" just below the question to do so.)

$\endgroup$
  • $\begingroup$ Does this look correct to you? $\endgroup$ – Filip Nov 25 '17 at 19:20
  • $\begingroup$ 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 $V$? $\endgroup$ – John Hughes Nov 25 '17 at 20:05
  • $\begingroup$ Do you mind giving me another hint? I'm kind of stuck? $\endgroup$ – Filip Nov 26 '17 at 10:08
  • $\begingroup$ 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. $\endgroup$ – John Hughes Nov 26 '17 at 13:12
  • $\begingroup$ 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? $\endgroup$ – Filip Nov 26 '17 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.