Characteristic polynomial has degree n and leading coefficent (-1)^n

[Bipolarity]

I've been looking for proof of the fact that the characteristic polynomial of an n by n matrix has degree n with leading coefficient ## (-1)^{n} ##.

I first tried proving it myself but my method is a bit strange (it does use induction though) and I am doubting the rigor, so could perhaps someone please post a link to the proof? I've been looking around without much success. Thanks!

BiP

 

[jgens]

If you want to do this by induction, then my suggestion would be to use the cofactor expansion of the determinant. There are other ways of doing this though that avoid induction.

 

[hilbert2]

could perhaps someone please post a link to the proof? I've been looking around without much success. Thanks!

http://www.math.uri.edu/~eaton/NotesCh5.pdf , see Theorem 5.3.

 

[Mandelbroth]

I've been looking for proof of the fact that the characteristic polynomial of an n by n matrix has degree n with leading coefficient ## (-1)^{n} ##.

I first tried proving it myself but my method is a bit strange (it does use induction though) and I am doubting the rigor, so could perhaps someone please post a link to the proof? I've been looking around without much success. Thanks!

BiP

Well, it seems fairly straightforward. You have a matrix ##\mathbf{A}=\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}\end{bmatrix}##.

To find the characteristic polynomial, we consider ##\det(\mathbf{A}-\lambda\mathbf{I})=\det\begin{bmatrix} a_{1,1}-\lambda & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2}-\lambda & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}-\lambda\end{bmatrix}=0##. The proof should then become fairly evident when considering the definition of the determinant. I'm tired and lazy right now, though, so I'm going to leave it at that.

As an aside, some authors choose to define the characteristic polynomial as ##\det(\lambda\mathbf{I}-\mathbf{A})=0##, in which case the factor of ##(-1)^n## is unnecessary.

I'd like to see your proof by induction, though. That might be a fun read.

 

[blue_raver22]

olarBear

The characteristic polynomial of an n by n matrix is defined as the polynomial with eigenvalues of the matrix as its roots. This polynomial has degree n because an n by n matrix has n eigenvalues. This can be seen by considering the fact that the characteristic polynomial is a polynomial of degree n and the fundamental theorem of algebra states that a polynomial of degree n has n complex roots. Since eigenvalues can be complex numbers, this means that an n by n matrix has n eigenvalues and therefore the characteristic polynomial has degree n.

The leading coefficient of the characteristic polynomial is determined by the determinant of the matrix. The determinant of an n by n matrix is defined as the sum of products of n elements of the matrix, where each product contains one element from each row and one element from each column. The sign of this sum is determined by the number of permutations of the elements, with a negative sign for an odd number of permutations. Since the leading term of the characteristic polynomial is the product of the eigenvalues, and the number of permutations is equal to the number of eigenvalues, the sign of the leading coefficient will be (-1)^n.

To prove this rigorously, one can use the definition of determinant and the properties of determinants to show that the sign of the leading coefficient is determined by the number of eigenvalues. This can be done through induction, as you have mentioned, or by using other methods such as the Leibniz formula for determinants.

In summary, the degree of the characteristic polynomial of an n by n matrix is n, and the leading coefficient is (-1)^n. This can be proven rigorously using the properties of determinants and the definition of eigenvalues. I hope this helps and feel free to ask for more clarification if needed.