Determining if the following sets are subspaces or not

linear algebravector-spaces

I was hoping someone could verify if my reasoning was correct on these questions to see if the following sets $(B_i)$ are subspaces of the corresponding vector space $(V_i)$ or not.

$V_1=P_2, B_1=\{p \in P_2|p+xp' \text{ has degree exactly 2}\}.\text{($p'$ is the he derivative of $p$).}$

I think this one IS a vector space.

the polynomial is essentially $ax^2+bx+c+x(2ax+b)$.

This implies that we have $ax^2+(b+2a)x+(c+b)$.

Now, I set $b+2a=\alpha$ and $c+b=\beta$.

So this way, I have $ax^2+\alpha x+\beta$.

Zero vector:

This clearly exists if I set $a=0$, $\alpha=0$ and $\beta=0$.

Closure under addition:

Suppose I have $ax^2+\alpha x+\beta$ and $dx^2+\gamma x+\theta$.

This would mean that $(ax^2+\alpha x+\beta)$+$(dx^2+\gamma x+\theta)$ gives $fx^2+gx+h$ if I define $a+d=f$, $\alpha + \gamma = g$ and $\beta + \theta = h$.

So we have closure under addition.

Closure under scalar multiplication:

Here, I take $ax^2+\alpha x+\beta$ and multiply by some constant $k$. This means I have:

$k(ax^2+\alpha x+\beta)$ or $lx^2+m x+n$ which means I have closure under scalar multiplication.

So $B_1$ is indeed a subspace.

$V_2 = M_2(\mathbb{R}), B_2=\{X \in M_2{\mathbb(R)|XA=0}, \text{where } A= \begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6
\end{pmatrix}$

Zero vector:

If I set the entries in $X$ to $0$, then $XA=0$ so thus, the zero vector is satisfied.

Closure under addition:

If I define an element $X_1$ and $X_2$, then $(X_1 + X_2)A=(X_1*A+X_2*A)=0+0=0$ so we have closure under addition.

Closure under scalar multiplication:

If I define a constant $c|c \in \mathbb{R}$, then we have $c(XA)$. However, $XA=0$, so we have $c(0)=0$ so we have closure under addition.

So $B_2$ is indeed a subspace.

$V_3=S, B_3=\{(a_n)_n \in S|a_n = a^2_{n-1} \text{ for all } n\geq 1\}.$

Zero vector:

If $n \geq 1$, then I define $a_0=0$ and $a_1=0$ so that I have $0=0^2$ or $0=0$. So the zero vector exists.

Closure under addition:

$a_n+b_n = a^2_{n-1} + b^2_{n-1}$ which implies that $c_n = c^2_{n-1}$. So the vector space is closed under scalar addition.

Closure under scalar multiplication:

$c*a_n = c*a^2_{n-1}$ implies $d_n = d^2_{n-1}$.

So we have closure under scalar multiplication.

Therefore, $B_3$ is a subspace.

$V_4 = F, B_4 = \{ f \in F|f(0)=f(1) \}$

This is one I am not sure about. I think the zero vector exists since I could just define $f(0)=0$ and $f(1)=0$ but I am not sure how I would show addition or scalar multiplication.

In a follow up, is this even a subspace in the first place? I can't tell.

If someone could verify if my reasoning is correct on the first three sets and explain the 4th set a little more, I would greatly appreciate it!

Best Answer

  • If $\mathbf 0(x)$ is the zero polynomial, then $\mathbf 0'(x) + x\mathbf 0(x) = 0$, and this has no degree exactly $2$.
  • Fine!
  • Closed under addition means that if $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are two sequences of real numbers such that $a_n = a_{n-1}^2$ and $b_n = b_{n-1}^2$ for all $n \geq 1$, then $a_n+b_n = (a_{n-1}+b_{n-1})^2$ for all $n \geq 1$. So...
  • If $\mathbf0$ is the zero function, then $\mathbf0(0) = 0 = \mathbf0(1)$, which means $\mathbf0 \in B_4$. Similarly, if $f,g \in B_4$ and $c$ is a real number, then $(cf+g)(0) = cf(0)+g(0) = cf(1)+g(1) = (cf+g)(1)$, which means $cf+g \in B_4$.
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