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I need help with an assignment, been stuck for two days with it. Any help/hint to (b), (d) or (e) would be greatly appreciated!

The linear transformation $T$: $\mathbb{R}^5 \longrightarrow \mathbb{R}^4$ is given by

$$ T \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 2x_1 - 4x_2 - x_3 - 3x_4 + 2x_5 \\ -x_1 + 2x_2 + x_3 + x_5 \\ x_1 - 2x_2 -x_3 - 3x_4 - x_5 \\ -x_1 + 4x_2 -x_3 + x_5 \\ \end{matrix}\right] , x = \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] \in \mathbb{R}^5 $$

(a) Decide the matrix $\mathbf{A}$ which fullfill $\mathbf{T(x) = Ax}$ for all $\mathbf{x} \in \mathbb{R}^5$.

(b) Let $y$ = ($y_1$ $y_2$ $y_3$ $y_4$)$^T \in \mathbb{R}^4$ be any (but unknown) vector.

Decide a vector $x \in \mathbb{R}^5$ (expressed by the unknowns $y_1, y_2, y_3, y_4$) which fullfull $T(x) = y$

(c) Decide a basis for kernel $T$. A linear transformation $S$: $\mathbb{R}^4 \longrightarrow \mathbb{R}^5$ which fullfill $(T \circ S)(y) = y$ for all $y \in \mathbb{R}^4$ is called a $rightinverse$ to $T$.

(d) Explain that any rightinverse to T is injective

(e) Find two different rightinverses, S, and S' to T.

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b) You're solving $Tx = y$ for $x$, in other words, $$\begin{bmatrix}2x_1 - 4x_2 - x_3 - 3x_4 + 2x_5 \\ -x_1 + 2x_2 + x_3 + x_5 \\ x_1 - 2x_2 - x_3 - 3x_4 - x_5 \\ -x_1 + 4x_2 - x_3 + x_5 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix}$$ for $x_1, \ldots, x_5$. Put this in an augmented matrix $$\left[\begin{array}{ccccc|c} 2 & 4 & -1 & -3 & 2 & y_1 \\ -1 & 2 & 1 & 0 & 1 & y_2 \\ 1 & -2 & -1 & -3 & -1 & y_3 \\ -1 & 4 & -1 & 0 & 1 & y_4 \end{array}\right]$$ Row reduce this matrix down to its reduced row-echelon form. I'll get you started: swap rows $1$ and $3$:

$$\left[\begin{array}{ccccc|c} 1 & -2 & -1 & -3 & -1 & y_3 \\ -1 & 2 & 1 & 0 & 1 & y_2 \\ 2 & 4 & -1 & -3 & 2 & y_1 \\ -1 & 4 & -1 & 0 & 1 & y_4 \end{array}\right]$$

then add row $1$ to row $2$:

$$\left[\begin{array}{ccccc|c} 1 & -2 & -1 & -3 & -1 & y_3 \\ 0 & 0 & 0 & -3 & 0 & y_2 + y_3 \\ 2 & 4 & -1 & -3 & 2 & y_1 \\ -1 & 4 & -1 & 0 & 1 & y_4 \end{array}\right].$$

Once it's in its reduced row-echelon form, you'll have a column without a leading $1$ (it may be $x_5$, it may not be). You may set this variable to be any number (or indeed, any function of $y_1, \ldots, y_4$) that you wish. I suggest setting it to $0$. Then, the remaining $x_i$ variables should come out to be linear functions of $y_1, \ldots, y_4$. This gives you a vector $x$ that produces $y$ when transformed under $T$ (and it doesn't hurt to verify this by transforming this proposed $x$ by $T$ and verify it cancels back to $y$).

d) This is a property of right inverses of any function, not just linear functions. To prove $S$ is injective, start with the assumption that $Sx = Sy$ for $x, y \in \Bbb{R}^5$. What happens if we apply $T$ to both sides?

e) The answer to (b) should have been a right inverse to $T$, provided you chose the free parameter to be $0$, or some linear function of $y_1, \ldots, y_4$. To get a second right inverse, instead set this free parameter to be a different function of $y_1, \ldots, y_4$, e.g. $y_1 + y_2$.

(You can also obtain non-linear right inverses by choosing non-linear functions of $y_1, \ldots, y_4$!)

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  • $\begingroup$ e) Why should the answer from b be a right inverse and not a left inverse or a general inverse? Is there any relationship that makes you able to say this beforehand? I mean it is easy to test, but some algebraic reasoning would be nicer. $\endgroup$ – Fac Pam May 17 '19 at 19:59
  • $\begingroup$ @FacPam: In part b), you're finding some function $S : \Bbb{R}^4 \to \Bbb{R}^5$, such that for all $y = (y_1, y_2, y_3, y_4)$, it maps to a vector $x = Sy$ with the property that $T(Sy) = Tx = y$. That is, $TS = I$. We can't have $ST = I$ too, as non-square matrices are never invertible. $\endgroup$ – Theo Bendit May 18 '19 at 1:23
  • $\begingroup$ @TheoBendit: Just wanted to say thank you for your hints. Made me realize a lot about the assignment. I had the hardest time with 1e, even with your hint, it felt more abstract. But really, thank you. Meant so much for me yesterday! $\endgroup$ – jubibanna May 18 '19 at 6:41
  • $\begingroup$ @jubibanna Great! I'm glad I could help. $\endgroup$ – Theo Bendit May 18 '19 at 8:34

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