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I want to check if the following matrix is linearly independent or not:

$$\begin{bmatrix}0&1\\0&1\end{bmatrix}$$

Now, this should be easy enough. This is what I do:

$$\lambda_1\begin{bmatrix}0\\0\end{bmatrix}+\lambda_2\begin{bmatrix}1 \\ 1\end{bmatrix} =0$$

If I find that $\lambda_1 =0$ and $\lambda_2=0$ are the only solutions to the system then the matrix is linearly independent, otherwise it is linearly dependent. I multiply the scalars by the vectors:

$$\begin{bmatrix}0\\0\end{bmatrix}+\begin{bmatrix}\lambda_2 \\ \lambda_2\end{bmatrix} =0$$

I get

$$\lambda_2=0$$ $$\lambda_2=0$$

So basically my solutions are $\lambda_2=0$ and $\lambda_1$ can be any number. Any hints if my reasoning is correct? I inserted the two vectors $v_1 = (0,0)$ and $v_2 = (1,1)$ of the matrix here and it says they're dependent. I'm not sure if I chose the right vectors though.

Please note my textbook assumes I don't know about ranks or determinants yet. So these are out of the equation. I should be able to determine if they are linearly dependent or independent by looking at the linear combination and doing something with it.

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If, when you say that a matrix is linearly dependent, what you mean is that its columns are linearly dependent, then you are right. I would add a concrete example of scalars $\lambda_1$ and $\lambda_2$ (not both equal to $0$) such that$$\lambda_1\begin{bmatrix}0\\0\end{bmatrix}+\lambda_2\begin{bmatrix}1 \\ 1\end{bmatrix} =0,$$such as, for instance, $\lambda_1=1$ and $\lambda_2=0$.

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  • $\begingroup$ So linear dependence means that at least one column can be expressed as a linear combination of the remaining columns. Could you please make an example of how somebody could be able to get say, the first column? I can't come up with an example of how this could be possible. $\endgroup$ – Cesare Oct 1 '18 at 12:47
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    $\begingroup$ @Cesare The first column is equal to $0$ times the second one, right?! $\endgroup$ – José Carlos Santos Oct 1 '18 at 12:51
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Your proof is correct.

The trivial solution is that the rows are identical which are obviously dependent but probably your instructor wants it done your way.

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I guess you mean are the columns (or rows) linearly independent. The answer is no.

One way is to compute the determinant. It's $0$.

Or, note that $1\cdot \begin{pmatrix} 0\\0\end{pmatrix}+0\cdot\begin{pmatrix}1\\1\end{pmatrix}=0$, is a nontrivial linear combination.

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