Linear Algebra and Vector spaces

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I am having a little trouble with the following questions.

Q: Are the following sets vector spaces with the indicated operations

a) The set of all polynomials of degree <=3 ; the operations of P

b) The set V of 2x2 matrices with zero determinant, usual matrix operations

c)The set V of all 2x2 matrices whose entries sum to zero; operations of M22

For part (a), I understand what polynomials with degree <= 3 means but I am not sure what the operations of P are supposed to be.

For part (b) I don't know what it means when a matrix has a determinant of zero but I do know what the usual matrix operations are

For part (c) I don't know what the operations of M22 are

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For a subspace all the following three properties must be satisfied:

$1) \ \vec{0} \in W\\ 2) \ \vec{v}+\vec{w} \in W\\ 3) \ \vec{cv}\to c \cdot \vec{v} \ ,c \in \mathbb{R}$

You have to check this properties and show that

For a)

The properties hold, indeed

1) Trivial

2) $(ax^3+bx^2+cx+d) +(ex^3+fx^2+gx+h) = (a+e)x^3+(b+f)x^2+(c+g)x+(d+h)$

3) $k(ax^3+bx^2+cx+d)=kax^3+kbx^2+kcx+kd$

For b) not, indeed

$\begin{pmatrix} 1 & 0 \\0 & 0\end{pmatrix}+\begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

For c) yes. 1) Trivial

2) Let consider

$M=\begin{pmatrix} a & b \\c & d\end{pmatrix}$ such that $a+b+c+d=0$

$N=\begin{pmatrix} e & f \\g & h\end{pmatrix}$ such that $e+f+g+h=0$

then

$M+N=\begin{pmatrix} a+e & b+f \\c+g & d+h\end{pmatrix}$ and $(a+e)+(b+f)+(c+g)+(d+h)=(a+b+c+d)+(e+f+g+h)=0$

3) $kM=\begin{pmatrix} ka & kb \\kc & kd\end{pmatrix}$ and $ka+kb+kc+kd=k(a+b+c+d)=0$

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It's always standard addition and scalar multiplication:

a) Sum of $2$ polynomials of degree $\leq 3$ is still a such polynomial and scaling such a polynomial does not increase its degree. Finally as the constant polynomial $0$ is of degree $\leq 3$ this is a vector space

b) You can have the sum of $2$ matrices with determinant $0$ to have a non-zero determinant (I'll let you think of an example yourself), hence this is not a vector space

Can you now do c) yourself? Hint: it is a vector space