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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.

Nażysław Zbyłutowicz
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2 Answers2

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

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.

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erfink
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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.

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