Direct sum of vector sub spaces in $\mathbb{R}^4$

Question

I am struggling to solve the below question:

$$ \ $$ let $V$ and $W$ be subspaces of $\mathbb{R}^4$: $$ V=\{(a,b,c,d) \in \mathbb{R}^4 \mid a+c=0,b+2d=0\} \\[2ex] W=\{(a,b,c,d) \in\Bbb R^4\mid a=2b,c=d\}$$

Need to check whether $\Bbb R^4=V\text{ Direct Sum }W$.

I approached it by taking inverse of system of equations as below: $$ \begin{cases} a+c=y_1 \\ b+2d=y_2 \\ a-2b=y_3 \\ c-d=y_4 \end{cases} $$

and able to get: $$ \begin{cases} a=\frac{4y_1 - 2y_2 - y_3 - 4y_4}{3} \\ b=\frac{2y_1 - y_2 - 2y_3 - 24y_4}{3} \\ c=\frac{y_1 - 2y_2 - y_3 - 4y_4}{3} \\ d=\frac{-y_1 + 2y_2 + y_3 + y_4}{3} \end{cases} $$

However I am not able to represent them as direct sum of two tupples say $v_1$ element of $V$ and $w_1$ element of $W$ where $v_1$ directSum $w_1$

This is not an assignment question.

Answer 1

Both subspaces are clearly two-dimensional. For $\mathbb R^4$ to be their direct sum, we must have $V\cap W=\{0\}$. Each of these subspaces is defined by a system of two homogeneous linear equations; their intersection is the solution space of the combined system of four equations. How many solutions does this system have?

Answer 2

It is clear that $$V=span\{\begin{pmatrix}-1\\ 0\\ 1\\ 0 \end{pmatrix},\begin{pmatrix}0\\ -2\\ 0\\ 1 \end{pmatrix}\}=span\{\textbf{v}_1,\textbf{v}_2\}$$ $$W=span\{\begin{pmatrix}2\\ 1\\ 0\\ 0 \end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\\ 1 \end{pmatrix}\}=span\{\textbf{w}_1,\textbf{w}_2\}$$

If every vector form $\mathbb{R}^{4}$ can be written as the linear combination v+w, where v $\in V$ and w $\in W$, then we are nearly done.

Let's check it now.

Since there are exactly 4 vectors from $\mathbb{R}^{4}$ above, we may use determinant to check whether those 4 vectors from a basis of $V+W$, span $\mathbb{R}^{4}$ or not .

As $$ \begin{vmatrix}-1 & 0 & 2 & 0\\ 0 & -2 & 1 & 0\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1 \end{vmatrix}=3 \neq 0$$, those 4 vectors are linearly independent and span $\mathbb{R}^{4}$.

It means $$\mathbb{R}^{4}=V+W$$

Now, the last thing we need to make sure is $$V\cap W =\{\textbf{0}\}$$.

Clearly, $\{\textbf{0}\} \subseteq\ V\cap W$ as $V\cap W$ is a subspace of $\mathbb{R}^{4}$.

To show $\{\textbf{0}\} \supseteq\ V\cap W$, picking any vector x $\in V\cap W $

$$\textbf{x}=c_1\textbf{v}_1+c_2\textbf{v}_2=d_1\textbf{w}_1+d_2\textbf{w}_2$$ $$\implies \textbf{0}=\textbf{x}-\textbf{x}=c_1\textbf{v}_1+c_2\textbf{v}_2-(d_1\textbf{w}_1+d_2\textbf{w}_2)$$

As these 4 vectors are linearly independent, the equation above has only trivial solution.

It yields the result $c_1=c_2=-d_1=-d_2=0$, so $\textbf{x}=\textbf{0}$.

Now we may conclude that $$V\cap W =\{\textbf{0}\}$$

So the sum $\mathbb{R}^{4}=V+W$ is a direct sum.

i.e. $\mathbb{R}^{4}=V \oplus W$