Artin Problems: Section 1.M

Problems from Chapter 1: Matrices, Miscellaneous Problems

More updates on this post to come.

Problem 1.M.1 :

Let a 2n \times 2n matrix be given in the form

M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}

where each block is an n \times n matrix. Suppose A is invertible and AC = CA. Use block multiplication to prove that 

det M = det (AD-CB). and give an example to show that this doesn’t hold if AC \neq CA

Problem 1.M.2 :

Let A be an m \times n matrix with m < n. Prove that A has no left inverse by comparing to A to the square n\times n matrix obtained by adding m-n rows of zeros to the bottom. 

Problem 1.M.3 :

The trace of a square matrix is the sum of it’s diagonal entries. Show the following properties of the trace.

(a) trace(A+B) = trace(A)+trace(B)

(b) trace(AB) = trace(BA)

(c) If B is invertible then trace(A) = trace(BAB^{-1})

trace(A+B) = \sum\limits_{i=1}^n a_{ii}+b_{ii} = \sum\limits_{i=1}^n a_{ii}+\sum\limits_{i=1}^n b_{ii} = trace(A)+trace(B) thus (a)

trace(AB) = \sum\limits_{i=1}^n ( \sum\limits_{j=1}^n a_{ij}\cdot b_{ji})= \sum\limits_{i=1}^n ( \sum\limits_{j=1}^n b_{ij}\cdot a_{ji})= trace(BA) therefore (b)

Problem 1.M.4 : Show that the equation AB-BA=I has no solution in real n\times n matrices AB.

We take the identity and use it to show a contradiction

AB-BA=I
I-B^{-1}A^{-1}BA=B^{-1}A^{-1}
I = B^{-1}A^{-1}+B^{-1}A^{-1}BA
I = B^{-1}A^{-1}(I+BA)

note that BA=I-AB

I = B^{-1}A^{-1}(2I-AB)
I = 2B^{-1}A^{-1}-I
2I = 2B^{-1}A^{-1}
I = B^{-1}A^{-1}

This implies that

B = A^{-1} \Leftrightarrow A = B^{-1}

It follows that

AB-BA = I \Rightarrow I-I = I = 0

A contradiction. Therefore no such A,B exist.

Problem 1.M.5 :

Problem 1.M.6 :

Problem 1.M.7 :

Problem 1.M.8 :

Problem 1.M.9 :

Problem 1.M.10 :

Problem 1.M.11 :

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