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Let $T_1$ and $T_2$ be two Linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$.

Let $\{x_1,x_2,\cdots,x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1(x_i)\neq 0$ for every $1\leq i\leq n$ and that $x_i\perp Ker (T_2)$ for every $1\leq i\leq n$.

Which of the following is true?

  • $T_1$ is invertible
  • $T_2$ is invertible
  • Both $T_1$ and $T_2$ are invertible
  • Neither $T_1$ nor $T_2$ is invertible.

As $T_1(x_i)\neq 0$ for each $1\leq i\leq n$ we do not have $T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 $ unless each $a_i=0$ i.e.,$a_1x_1+a_2x_2+\dots+a_nx_n=0$ i.e., $T_1$ is one one thus invertible.

I am not sure if $T_2$ is invertible or not.

we have $x_i\perp Ker (T_2)$ for all $x_i$. Would that be a good idea say something like

$\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)$ and as they span whole space we would have

$\mathbb{R}^n\perp Ker(T_2)$ and thus $Ker(T_2)=0$ so it is injective so it is invertible...

Thus both $T_1$ and $T_2$ are invertible?

I would be so thankful if someone can assure what it has been done here is sufficient/clear.

THank you

2 Answers2

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You are wrong that $T_1$ is invertible; it need not be. Indeed for $n=2$, $\{x_1,\ldots,x_n\}$ the standard basis, and $T_1$ given by the matrix $$\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$$ neither of $T_1(x_1)=x_1-x_2$ nor $T_1(x_2)=x_2-x_1$ is zero, but $T_1(x_1+x_2)=0$. Moreover one can take $T_2=I$ to satisfy $x_i\perp\ker(T_2)=\{0\}$ for $i=1,2$.

Your argument that $\ker(T_2)=\{0\}$ always, hence $T_2$ is invertible, is correct.

Marc van Leeuwen
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  • Oh yes... but how did my argument failed? :O –  Dec 26 '13 at 09:33
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    As you can check in the concrete examples I gave, it can happen that a sum of terms is $0$ without the individual terms being $0$. The argument you gave says that $T_1(a_1x_1+\cdots+a_nx_n)=a_1T_1(x_1)+\cdots+a_nT_1(x_n)$ cannot be zero unless all the $a_i$ are zero; this is just not true: it can be zero without that. – Marc van Leeuwen Dec 26 '13 at 15:23
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As defined, $T_1$ is not necessarily invertible. Consider the case for $n=2$. You may have $T(x_1)=(a,b)$ and $T(x_2)=(-a,-b)$, then $c_1=1; c_2=-1$ will give you $c_1T(x_1)+c_2T(x_2)=0$

For the second map $T_2$ , you can argue like this ( simplified, as suggested by Marc Van Leewen): if $Ker(T_2) \perp x_i$ for each $x_i$ in the basis, and $v_i$ is in $Ker(T_2)$then the set {$x_1,x_2,...,x_n,v_i$} is a linearly-independent set in $\mathbb R^n$ ( or in an n-dimensional vector space) , which is not possible. This forces $Ker(T_2)$={$0$} , so that $T_2$ is invertible.

user99680
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    I don't see the point of orthogonalizing your basis. Being a basis it spans everything, this is all that matters; orthogonalizing does not add anything useful. In fact after orthogonalizing you note that $n$ orthogonal vectors span the space, and any collection of orthogonal spaces (I suppose you meant vectors) are linearly independent, but all that was given for $x_1,\ldots,x_n$ at the outset! The argument in the question for this part is correct, and has nothing missing. – Marc van Leeuwen Dec 26 '13 at 07:36
  • Thanks; I corrected 'spaces' into 'vectors'. You're right; it may be overkill. I was trying to argue that if $v_i$ in $Ker(T_2)$ is orthogonal to each basis vector $x_i$ , then the set $x_1,x_2,..,x_n, v_i$ is a linearly-independent set, which cannot happen in n-dimensional spaces. This forces Ker$T^2=${$0$}. Still, your idea is better; I simplified as in your suggestion, thanks for the idea. – user99680 Dec 26 '13 at 07:43