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So I know how to do questions like this:

enter image description here

But this one is throwing me off a bit.

For the linear transformation $T\colon \mathbb{R}^3 \to \mathbb{R}^2$, where $T(x, y, z) = (x − 2y + z, 2x + y + z)$: (a) Find the rank of $T$. (b) Without finding the kernel of $T$, use the rank-nullity theorem to find the nullity of $T$

Is this right?

I think the equations represent this matrix:

\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 1 \end{bmatrix}

It's a 3x2 matrix which means it represents a transformation from $R^3$ to $R^2$

rank T:

\begin{bmatrix} 1 & -2 & 1 \\ 0 & 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & \frac{-1}{5} \end{bmatrix} \begin{bmatrix} 1 & 0 & \frac{3}{5} \\ 0 & 1 & \frac{-1}{5} \end{bmatrix}

rank is 2 nullity is 1.

Conceptually did I do this right? The equations that I'm given represent equations that I can then convert into a coefficient matrix which I can use to determine the missing variables right?

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  • $\begingroup$ You don’t need to do all that work. The two rows of the matrix are obviously linearly independent, so its rank is 2. $\endgroup$
    – amd
    Sep 17, 2018 at 7:01
  • $\begingroup$ Why is it so obvious? $\endgroup$
    – Jwan622
    Sep 17, 2018 at 15:14
  • $\begingroup$ When you only have two vectors, they are linearly dependent iff one is a scalar multiple of the other—they’re on the same line, which is what linear dependence means in its most basic form. $\endgroup$
    – amd
    Sep 17, 2018 at 22:34

2 Answers 2

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More precisely, the matrix $$ \left( \begin{matrix} 1 & -2 & 1 \\ 2 & 1 & 1 \end{matrix} \right) $$ is the matrix associated to $T$ with respect to the standard bases of $\mathbb{R}^3$ and $\mathbb{R}^2$. Without doing reduction, the rank of $T$ is given by the rank of one of the biggest submatrices with non-vanishing determinant. In your case there is a submatrix of rank $2$ with determinant non-zero (as gimusi is showing), so the rank of $T$ is $2$.

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Yes your way is correct, as an alternative since

$$\det\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}\neq 0$$

we can conclude that $\operatorname{rank}=2$ and then nullity is $3-2=1$.

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