Find $\lambda $ so the dimension of the vector subspace is 2.

Question

$\left\{a,b,c\right\}\in \mathbb{R}^3$ are linearly independent vectors.

Find the value of $\lambda $, so the dimension of the subspace generated by the vectors:

$2a-3b,\:\:\left(\lambda -1\right)b-2c,\:\:3c-a,\:\:\lambda c-b$ is 2.

So, if I understand this correctly the span of the given vectors should have 2 linearly independent vectors, so I construct the matrix:

$$A=\begin{pmatrix}2&0&-1&0\\ -3&\lambda -1&0&-1\\ 0&-2&3&\lambda \end{pmatrix}$$

And this matrix should have rankA = 2? And now I should just find a value for lambda that satisfies this condition? Is my logic correct?

Answer 1

I'd say column reduction is easier here, since there are already two linearly independent columns that do not involve $\lambda$ at all; the remaining columns must be linear combinations of them. Column-reduction of $A$ (starting with moving the second column to the end) gives $$ A'=\begin{pmatrix}2&0&0&0\\ -3&1&0&0\\ 0&-2&\lambda-2 &2\lambda-4\end{pmatrix} $$ Now it is clear that each of the last columns will only be a linear combination of the first two columns if it is zero, which just happens to occur for the same value of $\lambda$, namely $\lambda=2$; this is your answer.

Answer 2

As Michael said, by using row reduction:

$$\begin{pmatrix}2&0&-1&0\\ -3&\lambda -1&0&-1\\ 0&-2&3&\lambda \end{pmatrix}$$

$$ \iff \begin{pmatrix}2&0&-1&0\\ 0&\lambda -1&-\frac{3}{2}&-1\\ 0&-2&3&\lambda \end{pmatrix} $$

$$ \iff \begin{pmatrix}2&0&-1&0\\ 0&-2&3&\lambda\\0&\lambda -1&-\frac{3}{2}&-1 \end{pmatrix}$$

$$ \iff \begin{pmatrix}2&0&-1&0\\ 0&-2&3&\lambda\\0&0&\frac{3}{2}(\lambda-2)&\frac{1}{2}(\lambda+1)(\lambda-2) \end{pmatrix}$$

Notice that the first row and second row has non-zero pivotal element: $2$ and $-2$ in different position so the first row vector and second row vector is linearly independent wherever $\lambda$ gets arbitary value.

In the same logic, if $\lambda \neq 2$, these three row vector will be linearly independent. Therefore, the rank of A is 3.

If $\lambda = 2$, the third row vector is $ ( \ 0 \ 0\ 0\ 0 )$ which shows that these three row vectors are linearly dependent. By arguement above, $ rank A = 2 $