Question 4

4

(i) Note that we’re free to pick any real value for x_1. Then we have one equation to determine x_2, x_3, x_4 meaning that we can pick values for 2 of these arbitrarily, which determines the value of the last. So each vector of X has the form:

\begin{pmatrix}u \\ v \\ s \\ -v - s\end{pmatrix} = u \begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} + v \begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix} + s \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}

These are clearly linearly independent so a basis for X is:

\left\{\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}\right\}

with \mathrm{dim}(X) = 3.

(ii)

Note that we have:

x_1 + x_2 = 0

x_3 = 2x_4

Since we have one equation determining x_1 and x_2, we are free to pick the value of one arbitrary, then the other is determined. Similarly, for x_3 and x_4 we are free to pick the value of one arbitrarily then the other is already determined. So any vector in Y has the form:

\begin{pmatrix}\lambda \\ -\lambda \\ 2 \mu \\ \mu\end{pmatrix} = \lambda \begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix} + \mu \begin{pmatrix}0 \\ 0 \\ 2 \\ 1\end{pmatrix}

These are linearly independent so a basis for Y is:

\left\{\begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 2 \\ 1\end{pmatrix}\right\}

with \mathrm{dim}(Y) = 2.

(iii) For X \cap Y we have:

x_2 + x_3 + x_4 = 0

x_1 + x_2 = 0

x_3 - 2x_4 = 0

We only have one equation determining x_1 so we pick it arbitrarily, say u. Then x_2 = -u. Then we have:

x_3 + x_4 = u

x_3 = 2x_4

so that:

3x_4 = u

giving:

x_4 = \dfrac u 3

and then:

x_3 = \dfrac {2u} 3

So any vector in X \cap Y has the form:

u\begin{pmatrix}1 \\ -1 \\ \frac 2 3 \\ \frac 1 3\end{pmatrix}

So a basis for X \cap Y is:

\left\{\begin{pmatrix}1 \\ -1 \\ \frac 2 3 \\ \frac 1 3\end{pmatrix}\right\}

\mathrm{dim}(X \cap Y) = 1.

(iv)

Note that:

\mathrm{dim}(X + Y) = \mathrm{dim}(X) + \mathrm{dim}(Y) - \mathrm{dim}(X \cap Y) = 3 + 2 - 1 = 4

So the basis of X + Y consists of 4 linearly independent vectors in \mathbb R^4, so X + Y = \mathbb R^4, which has basis:

\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\\0\\\end{pmatrix}, \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}

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