Question

Let H be the set of all vectors of the form (a-5b, b-2a,b,a), where a and b are arbitrary real scalars. Show that H is a subspace of \(\mathbb{R}^4\) and find a basis for it.

\[\left( \begin{matrix} a-5b \\ b-2a \\ b \\ a \end{matrix} \right) =a\left( \begin{matrix} 1 \\ -2 \\ 0 \\ 1 \end{matrix} \right) + b\left( \begin{matrix} -5 \\ 1 \\ 1 \\ 0 \end{matrix} \right) \] so H is spanned by those two vectors and form the basis for H. But shouldn't the basis have 4 vectors since it has dimension 4?

Also, how do I show that H is a subspace of \(\mathbb{R}^4\)?

Answer 1

To show that you have a subspace, you apply the Subspace Test. If \(\mathbb{H}\) is your subspace, then you must show:
1) Show that \(\vec{0} \in \mathbb{H}\)
2) Show that for any \(\vec{x}\) and \(\vec{y}\) both \(\in\) \(\mathbb{H}\), that:
i) \(\vec{x}+\vec{y} \in \mathbb{H}\) (i.e \(\mathbb{H}\) is closed under addition)
ii) \(\beta\vec{x} \in \mathbb{H}\) (i.e. \(\mathbb{H}\) is closed under scalar multiplication, where \(\beta \in \mathbb{R}\))

I will rewrite your subspace like this:
\[\mathbb{H}= \left\{\left.\begin{bmatrix} c \\ d \\ b \\ a \end{bmatrix} \right|c=a-5b, d=b-2a \right\} \]

STEP 1): Is the 0 vector in H?
\[ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}\] So we verify
\(0=0-5(0)\) (equation c = a - 5b)
\(0 = 0\)

and
\(0 = 0 - 2(0)\) (equation d = b - 2a)
\(0 = 0\)

So indeed \(\vec{0} \in \mathbb{H}\).

STEP 2 i) Is H closed under addition?
Let \[ \vec{x}=\begin{bmatrix} c_1 \\ d_1 \\ a_1 \\ b_1 \end{bmatrix}, \vec{y}=\begin{bmatrix} c_2 \\ d_2 \\ a_2 \\ b_2 \end{bmatrix}\\
\text{then } \\
\begin{align}\vec{x}+\vec{y} &=\begin{bmatrix} c_1+c_2 \\ d_1+d_2 \\ a_1+a_2 \\ b_1+b_2 \end{bmatrix}\\
&= \begin{bmatrix} (a_1-5b_1)+(a_2-5b_2) \\ (b_1-2a_1) + (b_2-2a_2) \\ a_1+a_2 \\ b_1+b_2 \end{bmatrix} \\
&= \begin{bmatrix} \color{blue}{a_1+a_2} -5(\color{red}{b_1+b_2}) \\ \color{red}{b_1+b_2} -2(\color{blue}{a_1+a_2}) \\ \color{blue}{a_1+a_2} \\ \color{red}{b_1+b_2} \end{bmatrix} \end{align}
\]
So yes, \(\vec{x}+\vec{y} \in \mathbb{H}\)

STEP 2 ii) Is H closed under scalar multiplication?
\[
\begin{align}\beta\vec{x} &=\begin{bmatrix} \beta c_1\\ \beta d_1 \\ \beta a_1 \\ \beta b_1\end{bmatrix}\\
&= \begin{bmatrix} \beta (a_1-5b_1)\\ \beta (b_1-2a_1)\\ \beta a_1 \\ \beta b_1\end{bmatrix} \\
&= \begin{bmatrix} \color{blue}{\beta a_1} -5\color{red}{\beta b_1} \\ \color{red}{\beta b_1} -2\color{blue}{\beta a_1} \\ \color{blue}{\beta a_1} \\ \color{red}{\beta b_1} \end{bmatrix} \end{align}
\]
So yes, \(\beta \vec{x} \in \mathbb{H}\)

By the Subspace Test, \(\mathbb{H}\) is a subspace of \(\mathbb{R^4} \)

Answer 2

Hm I realize i switched the a and b in the rows 3 and 4 of the vectors... but everything should still hold.

Answer 3

For the basis... yes indeed two vectors won't span in \(\mathbb{R}^4\), but the question asked to find a basis, so you should construct vectrs such that you get a basis for \(\mathbb{R}^4\). Although, I am guessing they want you to use the vectors \( (1, -2, 0, 1)^T\) and \((-5, 1, 1, 0)^T\) to do this?

Answer 4

@kirbykirby Ah I was getting confused with showing it's closed under addition. But it makes sense.

For the basis I'm guessing it wants the bases for \(\mathbb{R}^4\) and not just a basis for \(H\) right? Either way we have to use those two vectors and find two more.

Answer 5

ok. I think it wants a basis for R^4 too but it's not too clear to be honest :o. But assuming that, the best way to proceed as follows:

Since a basis must span the vector space \(\mathbb{R}^4\) and be linearly independent, then consider the vectors:
\[ \vec{v_1} = \begin{bmatrix} 1 \\ -2 \\ 0 \\ 1 \end{bmatrix}, \vec{v_2} =\begin{bmatrix} -5 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \vec{e_1}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \vec{e_2}=\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \vec{e_3}=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \vec{e_4}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\]
The last 4 are the standard basis vectors. Then clearly \(\{ \vec{v_1}, \vec{v_2}, \vec{e_1}, \vec{e_2}, \vec{e_3}, \vec{e_4}\}\) spans \(\mathbb{R}^4\) since the 4 standard basis vectors are a basis for \(\mathbb{R^4}\). To find a linearly independent set, consider:
\[\vec{0}= c_1\vec{v_1}+c_2 \vec{v_2}+c_3\vec{e_1}+c_4 \vec{e_2}+c_5\vec{e_3}+c_6 \vec{e_4}\]
Row reducing the corresponding coefficient matrix will give:
\[\begin{bmatrix} 1&-5&1&0&0&0 \\ -2&1&0&1&0&0 \\ 0&1&0&0&1&0 \\ 1 &0&0&0&0&1\end{bmatrix}\sim \begin{bmatrix} 1&0&0&0&0&1 \\ 0&1&0&0&1&0 \\ 0&0&1&0&5&-1 \\ 0 &0&0&1&-1&2\end{bmatrix}\]

This implies that \(\{ \vec{v_1}, \vec{v_2}, \vec{e_1}, \vec{e_2}\}\) is a linearly independent set of vectors. ANd since \(\mathbb{R}^4\) can be spanned minimally with 4 vectors, this we can choose this set as a basis.

Answer 6

So generally you add the standard basis vectors, and row reduce to see which columns linearly independent?

Answer 7

From what I remember, this is the trick we usually used when we had to construct a basis using a given set of vectors

Answer 8

I remember learning that, just couldn't find it in my textbook and obviously had forgotten it haha.

Answer 9

Would it also be alright to find the orthogonal complement to H and it's basis.
Because \(H+H^\bot=\mathbb{R^4}\) in our case. So wouldn't the basis for \(H\) and the basis for \(H^\bot\) form a basis for \(\mathbb{R}^4\)?

Answer 10

Yes i am 99% sure that would be correct as well

Answer 11

I just tried it seems to be correct. The method you gave is far easier though.
Thanks for the help!

Answer 12

awesome :) yw!