Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.6, Problem 11

Question

Let $X$ be the vector space of all complex $n \times n$ matrices and define $T \colon X \to X$ by $Tx \colon= bx$, where $b \in X$ is fixed and $bx$ denotes the usual product of matrices. I know that $T$ is linear.

Under what conditions does $T^{-1}$ exist?

If $b$ is an invertible matrix, then of course $T^{-1}$ exists.

But does the existence of $T^{-1}$ necessarily imply the invertibility of the matrix $b$?

What condition(s) (other than invertibility), if any, should $b$ satisfy in order for $T^{-1}$ to exist?

Knowing this fact about square matrices might also be useful for this questions: If $AB = I$ then $BA = I$ — Martin Sleziak, Dec 14 '14 at 08:54

score 3 · Answer 1 · answered Dec 14 '14 at 06:38

3

If $T$ is invertible, there is some $x$ such that $bx = Tx = I$, where $I$ is the identity matrix (why?).

Hence?

answered Dec 14 '14 at 06:38

PhoemueX

36,211

But $T$ could well be invertible even if the identity matrix is not in the range of $T$, couldn't it? – Saaqib Mahmood Dec 14 '14 at 06:48
For a (single-valued) mapping to be invertible, it is sufficient for the mapping to be injective. Am I right? – Saaqib Mahmood Dec 14 '14 at 06:49
1

@SaaqibMahmuud It this is the case, you should explain in your posts what is your definition of invertible/inverse/$T^{-1}$. But the definition, which is used most frequently, means that only bijective maps are invertible. See: Inverse of a Function exists iff Function is bijective. For injective/surjective functions you can find at least one-sided inverse: Left inverse iff injective; right inverse iff surjective. – Martin Sleziak Dec 14 '14 at 08:33
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@SaaqibMahmuud Having said that, in your case $T\colon X\to X$ is a linear map and $X$ is finite-dimentional. For such maps, if $T$ is injective then it also is surjective (hence bijective). – Martin Sleziak Dec 14 '14 at 08:39

score 2 · Accepted Answer · answered Dec 14 '14 at 07:10

2

Suppose $T$ has inverse $T^{-1}$. Then $T$ is an injective map between finite dimensional vector spaces. The rank-nullity theorem then implies the map is surjective, so PhoemueX's argument may be applied.

answered Dec 14 '14 at 07:10

Potato

41,411

Ok, if one only means "injective" by "invertible", then one also has to apply this reasoning. One similar point: My proof above only shows that the matrix $b$ has a right inverse. But by similar arguments (which I wanted the OP to figure out himself), this suffices to conclude that $b$ is actually invertible. +1. – PhoemueX Dec 14 '14 at 08:49

Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.6, Problem 11

2 Answers2