Abstract Nonsense

Crushing one theorem at a time

Halmos Sections 37 and 38: Matrices and Matrices of Linear Transformations(Pt. IV)


Point of post: This is a continuation of this post.

Remark: For some strange reason the fourth (this one) and the fifth (the previous one) got mixed up in the order of posting. The number is correct, this is the fourth post in this sequence and the one preceding it the fifth.

16.

Problem: Decide which of the following matrices are invertible and find the inverses of the one that are.

a) \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}

b) \begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}

c) \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}

d) \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}

e) \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}

f) \begin{pmatrix}1 & 0 & 1\\ 1 & 0 & 1\\ 1 & 0 & 1\end{pmatrix}

g) \begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}

Proof:

It’s clear that if M is any matrix (in general) then if Mx=\bold{0} for some non-zero x\in F^n then M can’t be invertible.

a) This is invertible, since \begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix} serves as an inverse.

b) This is not invertible since \begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\begin{pmatrix}-1\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}

c) This isn’t since \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}

d) This is since \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}^2=I_2

e) This isn’t since \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 0\\1 & 0 & 0\end{pmatrix}\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}

f) This isn’t since \begin{pmatrix}1 & 0 & 1\\1 & 0 & 1\\1 & 0 & 1\end{pmatrix}\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}

g) This isn’t since  \begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}

17.

Problem: For which values of \alpha are the following matrices invertible? Find the inverses whenever possible.

a) \begin{pmatrix}\alpha & 1\\ 1 & 0\end{pmatrix}

b) \begin{pmatrix}1 & \alpha\\ 1 & 0\end{pmatrix}

c) \begin{pmatrix}1 & \alpha\\ 1 & \alpha\end{pmatrix}

d) \begin{pmatrix}1 & 1\\ 1 & \alpha\end{pmatrix}

Proof:

a) This is always invertible since \begin{pmatrix}0 & 1\\ 1 & -\alpha\end{pmatrix} serves as an inverse.

b) This is invertible precisely when \alpha\ne 0. Indeed, if \alpha=0 then \begin{pmatrix}1 & 0\\ 1 & 0\end{pmatrix}\begin{pmatrix}0\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}. But, if \alpha\ne 0 then \begin{pmatrix}0 & 1\\ \frac{1}{\alpha} & \frac{-1}{\alpha}\end{pmatrix} functions as an inverse.

c) This is never invertible. Indeed, \begin{pmatrix}1 & \alpha\\ 1 & \alpha\end{pmatrix}\begin{pmatrix}-\alpha\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}.

d) This is invertible precisely when \alpha\ne 1. Indeed, if \alpha=1 we’ve already proven in the last exercise that this matrix isn’t invertible, and if \alpha\ne 1 then \displaystyle \frac{1}{\alpha-1}\begin{pmatrix}\alpha & -1\\ -1 & 1\end{pmatrix} serves as an inverse.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

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December 19, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , ,

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