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This question already has an answer here:

In wikipedia "In linear algebra, the Cayley–Hamilton theorem (named after the mathematicians Arthur Cayley and William Rowan Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation." $\iff$ If $A$ is an $n\times n$ martix and $I_n$ is the $n\times n$ identity martix and $\text{let}$ $f(\lambda)=\text{det}(\lambda I_n-A)$ then we get $f(A)=0$.

In my book and wiki both use a order polynomial expression to prove it.

The question is: Why can't we just let $\lambda=A $ $\Rightarrow$ $\lambda I_n=AI_n=A$ $\Rightarrow$ $f(A)=\text{det}(A-A)=0$

Can anyone help me? Thanks

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marked as duplicate by Gerry Myerson, rschwieb, Pedro Tamaroff, P.., Thomas Andrews Apr 15 '13 at 13:34

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    $\begingroup$ because $\lambda$ is a number? and you are substituting a matrix where a number is? $\endgroup$ – Lost1 Apr 15 '13 at 12:58
  • $\begingroup$ yes!Lost is right!also \lambda I_n=AI_n=A is not correct .if x_\lambda be eigen value correspond with \lmbda then you will have \lambda I_n x_\lambda=Ax_\lambda $\endgroup$ – Somaye Apr 15 '13 at 13:09
  • $\begingroup$ See this Wikipedia section $\endgroup$ – Marc van Leeuwen Apr 15 '13 at 13:25
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If $$ A=\begin{pmatrix} a_{1,1}&a_{1,2}&a_{1,3}&\ldots& a_{1,n-2}& a_{1,n-1}& a_{1,n}\\ a_{2,1}&a_{2,2}&a_{2,3}&\ldots& a_{2,n-2}& a_{2,n-1}& a_{2,n}\\ a_{3,1}&a_{3,2}&a_{3,3}&\ldots& a_{3,n-2}& a_{3,n-1}& a_{3,n}\\ \vdots&\vdots&\vdots&\ddots& \vdots& \vdots& \vdots\\ a_{n-2,1}&a_{n-2,2}&a_{n-2,3}&\ldots& a_{n-2,n-2}& a_{n-2,n-1}& a_{n-2,n}\\ a_{n-1,1}&a_{n-1,2}&a_{n-1,3}&\ldots& a_{n-1,n-2}& a_{n-1,n-1}& a_{n-1,n}\\ a_{n,1}&a_{n,2}&a_{n,3}&\ldots& a_{n,n-2}& a_{n,n-1}& a_{n,n}\\ \end{pmatrix} $$ then $$ \lambda I-A=\begin{pmatrix} \lambda -a_{1,1}&-a_{1,2}&-a_{1,3}&\ldots& -a_{1,n}& -a_{1,n-1}& -a_{1,n}\\ -a_{2,1}&\lambda -a_{2,2}&-a_{2,3}&\ldots& -a_{2,n-2}& -a_{2,n-1}& -a_{2,n}\\ -a_{3,1}&-a_{3,2}&\lambda -a_{3,3}&\ldots& -a_{3,n-2}& -a_{3,n-1}& -a_{3,n}\\ \vdots&\vdots&\vdots&\ddots& \vdots& \vdots& \vdots\\ -a_{n-2,1}&-a_{n-2,2}&-a_{n-2,3}&\ldots&\lambda -a_{n-2,n-2}& -a_{n-2,n-1}& -a_{n-2,n}\\ -a_{n-1,1}&-a_{n-1,2}&-a_{n-1,3}&\ldots& -a_{n-1,n-2}&\lambda -a_{n-1,n-1}& -a_{n-1,n}\\ -a_{n,1}&-a_{n,2}&-a_{n,3}&\ldots& -a_{n,n-2}& -a_{n,n-1}&\lambda -a_{n,n}\\ \end{pmatrix} \Longrightarrow\\ 0=A I-A=\begin{pmatrix} A -a_{1,1}&-a_{1,2}&-a_{1,3}&\ldots& -a_{1,n}& -a_{1,n-1}& -a_{1,n}\\ -a_{2,1}&A -a_{2,2}&-a_{2,3}&\ldots& -a_{2,n-2}& -a_{2,n-1}& -a_{2,n}\\ -a_{3,1}&-a_{3,2}&A -a_{3,3}&\ldots& -a_{3,n-2}& -a_{3,n-1}& -a_{3,n}\\ \vdots&\vdots&\vdots&\ddots& \vdots& \vdots& \vdots\\ -a_{n-2,1}&-a_{n-2,2}&-a_{n-2,3}&\ldots&A -a_{n-2,n-2}& -a_{n-2,n-1}& -a_{n-2,n}\\ -a_{n-1,1}&-a_{n-1,2}&-a_{n-1,3}&\ldots& -a_{n-1,n-2}&A -a_{n-1,n-1}& -a_{n-1,n}\\ -a_{n,1}&-a_{n,2}&-a_{n,3}&\ldots& -a_{n,n-2}& -a_{n,n-1}&A -a_{n,n}\\ \end{pmatrix} \\ ??? $$

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  • $\begingroup$ What a big mistake i have taken.......Thankyou $\endgroup$ – Xiaolang Apr 15 '13 at 13:07

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