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rossheed
 Post subject: Proof by Induction: Determinant of scalar times matrix
PostPosted: Sun, 19 Sep 2010 07:11:49 UTC 
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Joined: Sun, 27 Jun 2010 20:50:11 UTC
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Hi Guys,

My instructor was unable to complete any inductive proofs for linear algebra on the board, he simply said if it's true for k, it clearly must be true for k+1 (writing down nothing) - leaving us scratching our heads.

I have no experience with inductive proofs.

In homework I'm now perplexed on solving Let A \in \mathbb{C}^{n \times n}, \alpha \in \mathbb{C}, Show that \det(\alpha A) = \alpha^{n} \cdot \det(A)

I know the basis for induction is for n=1, \det(\alpha A) = \alpha \cdot a_{1 1} = \alpha \det(A)

And somehow I need to show that for n>1, \det(\alpha A) = \displaystyle\sum\limits_{i=1}^n a_{i 1} (-1)^{i+1} \det(\alpha M_{i 1} ) = \alpha^{n} \det(A)

Not sure how to go about this, any help is greatly appreciated.
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outermeasure
 Post subject: Re: Proof by Induction: Determinant of scalar times matrix
PostPosted: Sun, 19 Sep 2010 07:28:15 UTC 
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rossheed wrote:
Hi Guys,

My instructor was unable to complete any inductive proofs for linear algebra on the board, he simply said if it's true for k, it clearly must be true for k+1 (writing down nothing) - leaving us scratching our heads.

I have no experience with inductive proofs.

In homework I'm now perplexed on solving Let A \in \mathbb{C}^{n \times n}, \alpha \in \mathbb{C}, Show that \det(\alpha A) = \alpha^{n} \cdot \det(A)

I know the basis for induction is for n=1, \det(\alpha A) = \alpha \cdot a_{1 1} = \alpha \det(A)

And somehow I need to show that for n>1, \det(\alpha A) = \displaystyle\sum\limits_{i=1}^n a_{i 1} (-1)^{i+1} \det(\alpha M_{i 1} ) = \alpha^{n} \det(A)

Not sure how to go about this, any help is greatly appreciated.


Yuck, that is the wrong way to define determinant if you want to prove things.

Hint: what is M_{i1}? So what is \det(\alpha M_{i1})? And it should be \det(\alpha A)=\sum (-1)^{i+1}(\alpha A)_{i1}\det(\alpha M_{i1}).

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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rossheed
 Post subject: Determinant of scalar times matrix proved, I think
PostPosted: Sun, 19 Sep 2010 20:13:13 UTC 
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Joined: Sun, 27 Jun 2010 20:50:11 UTC
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Location: New Mexico, USA
Thanks outermeasure - I think I get the idea of proof by induction now. I'm changing the basis to a 2x2 since a 1x1 seems to be a trivial case, and gives little information on how the problem scales. That hint you gave me about writing it properly (because I had forgot to include alpha in the determinant equation) sparked my brain to figure out the rest. Many thanks and respect!

My proof:
Spoiler:
Basis:

For n=2, \det( \alpha A ) = \displaystyle\left| \begin{array}{cc} \alpha a_{1 1} & \alpha a_{1 2} \\ \alpha a_{2 1} & \alpha a_{2 2} \end{array} \right| = \alpha^{2} a_{1 1} a_{2 2} - \alpha^{2} a_{1 2} a_{2 1} = \alpha^{2} (a_{1 1}a_{2 2} - a_{1 2}a_{2 1} )~=~\alpha^{2}~\det(A)

For n=k, B \in \mathbb{C}^{k \times k}, and \alpha \in \mathbb{C} we assume, det(\alpha B) = \alpha^{k}det(B) is true.

For n=k+1, A \in \mathbb{C}^{k+1 \times k+1}, \alpha \in \mathbb{C}, det(\alpha A) = \displaystyle\sum^{k+1}_{j=1}\alpha a_{1 j} (-1)^{1+j} \det(\alpha M_{1 j} )

But since M_{1 j} \in \mathbb{C}^{k \times k}, \det( \alpha M_{1 j} ) = \alpha^{k} \det(M_{1 j}) by the n=k statement above.

\det(\alpha A) = \displaystyle\sum^{k+1}_{j=1} \alpha a_{1 j} (-1)^{1+j} \det(\alpha M_{1 j}) = \displaystyle\sum^{k+1}_{j=1} \alpha a_{1 j} (-1)^{1+j} \alpha^{k} \det(M_{1 j})

Factoring out \alpha, \det(\alpha A) = \alpha^{k+1} \displaystyle\sum^{k+1}_{j=1} a_{1 j} (-1)^{1+j} \det(M_{1 j}) = \alpha^{k+1} \det(A)
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