1. ## Linear maps

i have 2 questions im finding hard to try and answer.

Let T1 : U -> V and T2 : V -> W be linear maps.
Show that the map T2 T1 : U ->W is linear.

and

A matrix which satisfies A^T = -A is called skew-symmetric. Show that if A is an n*n skew-symmetric matrix where n is an odd integer, then A has no inverse. (Hint: use the definition A^T = -A to find det A, ) but be careful!

2. Hi
Originally Posted by sterps
i have 2 questions im finding hard to try and answer.

Let T1 : U -> V and T2 : V -> W be linear maps.
Show that the map T2 T1 : U ->W is linear.
Let $\displaystyle \lambda \in \mathbb{K},\,u_1,\,u_2\in U$, you need to show that $\displaystyle T_2\circ T_1(\lambda u_1+u_2)=\lambda T_2\circ T_1(u_1)+T_2\circ T_1(u_2)$ acknowledged that $\displaystyle T_1$ and $\displaystyle T_2$ are both linear maps.
A matrix which satisfies A^T = -A is called skew-symmetric. Show that if A is an n*n skew-symmetric matrix where n is an odd integer, then A has no inverse. (Hint: use the definition A^T = -A to find det A, ) but be careful!
You got the hint... (and I even think that the solution has been posted in this forum few days ago)

3. Originally Posted by sterps
i have 2 questions im finding hard to try and answer.

Let T1 : U -> V and T2 : V -> W be linear maps.
Show that the map T2 T1 : U ->W is linear.
and
A matrix which satisfies A^T = -A is called skew-symmetric. Show that if A is an n*n skew-symmetric matrix where n is an odd integer, then A has no inverse. (Hint: use the definition A^T = -A to find det A, ) but be careful!
1)If $\displaystyle T_1$ and $\displaystyle T_2$ are linear maps

Let $\displaystyle u,v \in U$, then $\displaystyle T_1(u+v) = T_1(u) + T_1(v) \in V$
Since $\displaystyle T_1(u),T_1(v) \in V$, then $\displaystyle T_2(T_1(u+v))) = T_2(T_1(u) + T_1(v)) = T_2(T_1(u)) + T_2(T_1(v)) \in W$

Thus $\displaystyle \forall u,v \in U$, $\displaystyle T_2(T_1(u+v))) = T_2(T_1(u)) + T_2(T_1(v))\in W$

Do similarly for scalar multiplication, thus proving that the map T2 T1 : U ->W is linear.

2) Claim: $\displaystyle det(A) = 0$

$\displaystyle A^T = -A$

$\displaystyle \Rightarrow det(A^T) = det(-A)$

$\displaystyle \Rightarrow det(A) = (-1)^n \,\, det(A) [\because \forall A_{n \times n}, \,\, det(\lambda A) = \lambda ^n\,det(A) \text{and } det(A^T) = det(A)]$

$\displaystyle \Rightarrow det(A) = -det(A)[\because \text{n = odd}]$

$\displaystyle \Rightarrow det(A) = 0$