# Explain that any rightinverse to $T$: $\mathbb{R}^5 \longrightarrow \mathbb{R}^4$ is injective & Find two different rightinverses, S, and S' to T. [closed]

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.

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 $$y_1, \ldots, y_4$$!)

• 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. – Fac Pam May 17 '19 at 19:59
• @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. – Theo Bendit May 18 '19 at 1:23
• @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! – jubibanna May 18 '19 at 6:41
• @jubibanna Great! I'm glad I could help. – Theo Bendit May 18 '19 at 8:34