# Finding the Kernel of a Matrix Map

• I
Gold Member
Hi All,
I am trying to see if there is a "nice" ( relatively straightforward) way of finding the
solution/kernel of the map : ##f(A)=A^n -Id ## , where A is an ## k \times k ## matrix and ##n## is a positive
integer. Question: what is the kernel of this map? Cranking out matrix coefficients gets messy
and very slow as n grows. I can think of a solution using the Fundamental Theorem of Linear
Algebra relating Ortho spaces of rows, columns and row- and column- spaces. Is there some other
"reasonable" way of finding all solutions (clearly Id,-Id are solutions for n even and Id is a solution for n odd
)?

StoneTemplePython
Gold Member
some more details would be of interest. Since you mention orthogonality, I'm assuming your scalars are in ##\mathbb R## or ##\mathbb C##.

What's the motivation here-- are you trying to calculate a (finite) von neumann series?

To confirm: is ##Id## the identity matrix ##I##? If so, the kernel is really a question of whether or not ##\mathbf A## has an eigenvalue with magnitude 1 (and is an nth root of unity), and what the geometric multiplicities of associated eigenvectors are...

another thought: if you want to go the direct multiplication route there is the approach of first finding ##\mathbf A^2## then squaring that to get ##\mathbf A^4## then squaring that to get ##\mathbf A^8## then squaring that to get ##\mathbf A^{16}##, and so on until you get 'close' to ##\mathbf A^n## then cleverly finishing off the residual... as I recall this cuts down multiplications to something like ##O\big(log(n)\big)##,in a manner similar to bisection.

• WWGD
Gold Member
some more details would be of interest. Since you mention orthogonality, I'm assuming your scalars are in ##\mathbb R## or ##\mathbb C##.

What's the motivation here-- are you trying to calculate a (finite) von neumann series?

To confirm: is ##Id## the identity matrix ##I##? If so, the kernel is really a question of whether or not ##\mathbf A## has an eigenvalue with magnitude 1 (and is an nth root of unity), and what the geometric multiplicities of associated eigenvectors are...

another thought: if you want to go the direct multiplication route there is the approach of first finding ##\mathbf A^2## then squaring that to get ##\mathbf A^4## then squaring that to get ##\mathbf A^8## then squaring that to get ##\mathbf A^{16}##, and so on until you get 'close' to ##\mathbf A^n## then cleverly finishing off the residual... as I recall this cuts down multiplications to something like ##O\big(log(n)\big)##,in a manner similar to bisection.
Yes, thanks, good points, yes, we would need a notion of orthogonality to make appeal to Fundamental Theorem, but I think it applies only to Real-valued matrices. Still, rotation by ## 2\pi/n ## is a solution , despite having no Real eigenvalues.

In the class of complex matrices all the solutions of the equation ##A^n=I## are given by ##A=SDS^{-1}##, where ##S## is an invertible matrix, and ##D## is a diagonal matrix whose diagonal entries are ##n##th roots of unity, ##d_{j}^n=1##.

In the class of real matrices all the solutions are given by ##A=SRS^{-1}##, where ##S## is a real invertible matrix, and ##R## is a "rotation" matrix, i.e. ##R## is a block diagonal matrix whose diagonal blocks can be either rotations through the angle ##2\pi j/n##, ##j=0, 1, 2, \ldots, n-1## (these are ##2\times 2## blocks) or just real roots of unity (##1\times 1## blocks). The real roots of unity are ##\pm 1## if ##n## is even, and ##1## if ##n## is odd.

A rotation through the angle ##\alpha## matrix is given by $$\left( \begin{array}{cc} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right).$$

Gold Member
In the class of complex matrices all the solutions of the equation ##A^n=I## are given by ##A=SDS^{-1}##, where ##S## is an invertible matrix, and ##D## is a diagonal matrix whose diagonal entries are ##n##th roots of unity, ##d_{j}^n=1##.

In the class of real matrices all the solutions are given by ##A=SRS^{-1}##, where ##S## is a real invertible matrix, and ##R## is a "rotation" matrix, i.e. ##R## is a block diagonal matrix whose diagonal blocks can be either rotations through the angle ##2\pi j/n##, ##j=0, 1, 2, \ldots, n-1## (these are ##2\times 2## blocks) or just real roots of unity (##1\times 1## blocks). The real roots of unity are ##\pm 1## if ##n## is even, and ##1## if ##n## is odd.

A rotation through the angle ##\alpha## matrix is given by $$\left( \begin{array}{cc} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right).$$
Thanks, Hawkeye, are all permutation matrices then also rotation matrices? The antidiagonal matrix ## A_{ij} ; a_{ii}=0 ; a_{ij}=-1 ; i \neq j ## satisfies ##A^2=Id ## but it cannot be a rotation matrix since otherwise ## sin \alpha =-sin \alpha =-1 ##. Did I make a mistake?EDIT: Or is this matrix conjugate to a rotation? EDIT2: I don't think it can be conjugate to a rotation matrix ; its determinant would still need to equal 1.

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Thanks, Hawkeye, are all permutation matrices then also rotation matrices? The antidiagonal matrix ## A_{ij} ; a_{ii}=0 ; a_{ij}=-1 ; i \neq j ## satisfies ##A^2=Id ## but it cannot be a rotation matrix since otherwise ## sin \alpha =-sin \alpha =-1 ##. Did I make a mistake?EDIT: Or is this matrix conjugate to a rotation? EDIT2: I don't think it can be conjugate to a rotation matrix ; its determinant would still need to equal 1.

I believe you are talking here about a ##2\times 2## matrix. Then your antidiagonal matrix is not a rotation matrix, but it is a matrix of form ##SDS^{-1}##, where $$D=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$

Gold Member
I believe you are talking here about a ##2\times 2## matrix. Then your antidiagonal matrix is not a rotation matrix, but it is a matrix of form ##SDS^{-1}##, where $$D=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$
Right, but I thought all solutions to ##A^n =Id ## consisted of rotations ( up to conjugation) , and the matrix I mentioned, antidiagonals =-1; 0 otherwise, is a solution (specifically, to ## A^2=Id ##), but not a rotation? But, I agree, if A is a rotation then its eigenvalues must be n-th roots of unity .

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Let me elaborate. In the class of real matrices, all the solutions are given by ##A=SRS^{-1}##, where ##R## is a "rotation" (note the quotes, that is not always a real rotation), and ##S## is invertible. "Rotation" here mean that ##R## is the block diagonal matrices, where each block is either a ##2\times 2## rotation matrix (through an allowed angle), or just ##1\times 1## block, with ##n##th root of unity in it. I probably did not make it clear, but you do not have to have both types of blocks: your matrix can have only ##2\times2## blocks, or only ##1\times1## blocks, or any combination of these 2 (that adds up to the dimension).

The matrix $$R=D=\left(\begin{array}{cc} 1 & 0\\ 0& -1\end{array}\right)$$ is of that type, and it is not a rotation (that why I used quotes). But your matrix can be represented as ##SDS^{-1}##, so your example does not contradict my statement.

• WWGD
StoneTemplePython
Gold Member
I think you two may be talking past each other.

Right, but I thought all solutions to ##A^n =Id ## consisted of rotations

This is wrong. For example: ##A^2 = I ## is an involution. Such a matrix does not need to be normal. (i.e. it is diagonalizable, but need not be unitarily diagonalizable.)

Permutation matrices are a special kind of orthogonal matrix (which is a special kind of unitary matrix). Rotation matrices are also orthogonal matrices. All orthogonal matrices are normal.

Another way to think about it is: Pick your favorite orthogonal matrix ##P##, where ##P^n = I##. Now do a non-unitary similarity transform, i.e. consider ##S## where ##S^{-1} \neq S^*##.

Now take a look at ##\big(S^{-1}PS\big)##. The spectrum of this is the same as for ##P## because the matrices are similar. But the ##P## is normal while the new matrix is not.

yet we still have

##P^n = I##

so

##\big(S^{-1}PS\big)^n =\big(S^{-1}P^n S\big) = \big(S^{-1}IS\big) = \big(S^{-1}S\big)= I ##

• WWGD
Gold Member
I think you two may be talking past each other.

This is wrong. For example: ##A^2 = I ## is an involution. Such a matrix does not need to be normal. (i.e. it is diagonalizable, but need not be unitarily diagonalizable.)

Permutation matrices are a special kind of orthogonal matrix (which is a special kind of unitary matrix). Rotation matrices are also orthogonal matrices. All orthogonal matrices are normal.

Another way to think about it is: Pick your favorite orthogonal matrix ##P##, where ##P^n = I##. Now do a non-unitary similarity transform, i.e. consider ##S## where ##S^{-1} \neq S^*##.

Now take a look at ##\big(S^{-1}PS\big)##. The spectrum of this is the same as for ##P## because the matrices are similar. But the ##P## is normal while the new matrix is not.

yet we still have

##P^n = I##

so

##\big(S^{-1}PS\big)^n =\big(S^{-1}P^n S\big) = \big(S^{-1}IS\big) = \big(S^{-1}S\big)= I ##
Thanks, I am aware not all solutions are rotations , by a straightforward counting argument, but I thought someone here had stated they are rotations, up to similarity.

Thanks, I am aware not all solutions are rotations , by a straightforward counting argument, but I thought someone here had stated they are rotations, up to similarity.

I used "rotation" in quotes for the lack of better term, it is not always a real rotation. Your example is up to similarity what I call "rotation" see post # 8 above for the details.