are there any matrices that act like the identity matrix but are totally different?

Given any matrix $A$ or vector $v$, can you find a matrix $B \neq I$ and also not resembling $I$ (having entries in places other than diagonal) such that $B*A$ and $B*v$ are approximately $A$ and approximately $v$?

For me the answer is should obviously be yes. Also I am aware that $I$ is unique, so the matrix $B$ will never have the same effect as $I$ unless it is equal to $I$ itself.

• What about just $\begin{bmatrix} 1.0001 & 0 \\ 0 & 1.0001 \end{bmatrix}$ ? Mar 30 '17 at 3:20
• Yes I have thought of this example but I would like the matrix to not resemble $I$. I have edited my post also. Mar 30 '17 at 3:20
• Maybe the question is are only diagonal matrices that resemble $I$ able to preserve the "spirit" of any given matrix $A$? Mar 30 '17 at 3:22
• I don't see why you think the answer should be yes. I think intuitively, the answer should be no. Especially if there are entries in other places than the trace, then it will have very different effects for different vectors. Mar 30 '17 at 3:29
• How about a rotation through a very small angle?
– amd
Mar 30 '17 at 4:47

$$BA \approx A \implies BA-A \approx 0 \implies (B-I)A \approx 0$$

If this is to hold for any matrix $$A$$, then we would need that $$B-I \approx 0 \implies B \approx I$$. What we mean by "$$\approx$$" would require a precise choice of norms / topology on the space of matrices, but would generally require that $$B= I + \epsilon C$$ where $$\epsilon$$ is "small" and $$C$$ is any matrix with entries that aren't "too big." For example, $$B= I + 0.001 \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix}$$. If you place more restrictions on $$B$$, then it will have to resemble the identity matrix even more closely.

Edit:

Due to some continued interest in this question, I would like to expound on some of the special properties that the identity matrix has which we might want $$B \approx I$$ to emulate. As above, we will always require $$AB \approx A$$ in addition to the considered property. Furthermore, other matrices ($$C$$, $$D$$, $$P$$, $$N$$, $$\dots$$) should all be considered matrices with entries that aren't "too big" in addition to other required properties.

1. $$I$$ commutes with any square matrix: $$AI = A = IA$$. In general, two matrices commute ($$AB=BA$$) if and only if "they respect each other's eigenspaces." If this is to happen for any matrix $$A$$, then $$B= I + \epsilon (d I)$$, where $$\epsilon$$ is "small" and $$dI$$ is a diagonal matrix with a repeated entry that isn't "too big", e.g., $$B= I + 0.001 \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$. This is because every vector must be an eigenvector for $$B$$.
2. If we relax (1.) to need only "approximate commutivity" of $$AB \approx BA$$, then we can relax back to $$B = I + \epsilon C$$ where $$C$$ is any matrix with entries that aren't "too big."
3. $$I$$ is a rotation by $$0$$ degrees in every direction, i.e., the identity matrix preserves the direction of every vector. This would require $$B= I + \epsilon (d I)$$.
4. If $$B$$ is to be "approximately irrotational" and fix lengths, then $$B=Q$$ where $$Q$$ is a unitary (orthogonal) matrix such that all (possibly complex) eigenvalues $$\lambda_i$$ of $$Q$$ satisfy $$| \lambda_i -1 |$$ being small and $$|\lambda_i| = 1$$.
5. $$I$$ preserves the row space, column space, null space, and left null space of $$A$$. This would require $$B= I + \epsilon (d I)$$.
6. $$I$$ is expanding (not strictly contracting): $$\| I \vec{v} \| \geq \| \vec{v}\|$$ for all $$\vec{v}$$. This would require $$B= I + \epsilon D$$ where $$D$$ has eigenvalues that are all greater than or equal to zero.
7. $$I$$ is contracting (not strictly expanding): $$\| I \vec{v} \| \leq \| \vec{v} \|$$. This would require $$B = I + \epsilon D$$ where the eigenvalues of $$D$$ are all less than or equal to zero.

If we look at the singular value decomposition, $$I=U \Sigma V^T$$ where $$U$$ and $$V$$ are rotations (by zero degrees) and $$\Sigma$$ is a stretching by the singular values $$\sigma_1= \sigma_2 = \dots = \sigma_n = 1$$. This can give a complete answer to any of the above cases. In general, $$B= \tilde{ U} \tilde{\Sigma} \tilde{ V}^T$$ where $$\tilde{U}$$ and $$\tilde{V}$$ are unitary (orthogonal) matrices that give "small rotations" and $$\tilde{\Sigma} = I + \epsilon D$$ where $$D$$ is a diagonal matrix.

• +1 for nicely sidestepping having to ask the OP what "resembles" means. Mar 30 '17 at 3:28

Yes you can, but for non-square matrix cases. The answer: get the hat matrix!

Suppose $$X$$ is a non-square matrix ($$n\times m$$) where $$n\neq m$$, then: $$Z = X(X'X)^{-1}X'$$ is the hat matrix ($$n\times n$$), which can be easily shown to have the property: $$ZX=X$$

Note if $$X$$ is square, $$Z$$ is the identity matrix.