Sometimes formulas in linear algebra are not easy to remember. Some usefulness for the process of remembering can provide application of mnemonics.

  • Do you know some useful mnemonics for this purpose?

I'll give two examples:

  • For the process of finding the inverse of matrix it could be used mnemonic Detminstra what can be translated as 1. calculate determinant 2. for every entry find minors with the sign 3. transpose obtained matrix.
  • The other example is Avvedia what is the shortcut for the formula with eigenvectors $AV=VD$. Knowing this formula we can easily obtain formula $A=VDV^{-1}$ or twin formula for diagonal matrix $D=V^{-1}AV$- sometimes $V$ and $V^{-1}$ can be erroneously interchanged - with Avvedia it is easier to check correctness of a formula.

What other mnemonics could be useful in linear algebra?

Added lately

  • Furorri: concerning existence of right inverse for full row rank matrix (analogously for full column rank matrix would be Fucorlin) - these two inverses easy to erroneously interchange.
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    $\begingroup$ If you've internalized the fact that $A \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} = \begin{bmatrix} A v_1 & \cdots & A v_n \end{bmatrix}$, then the formula $AV = VD$ is easy to remember because it tells us that $A v_i = d_i v_i$ for all $i$. Gilbert Strang emphasizes various useful ways of looking at matrix multiplication in his linear algebra books. If Cramer's rule is forgotten, it can be derived quickly using the approach here: math.stackexchange.com/a/1941606/40119 I could be wrong but I'm a bit skeptical of mnemonics in math education; it seems better to focus on deriving. $\endgroup$
    – littleO
    Mar 28, 2017 at 8:46
  • $\begingroup$ @littleO However deriving is always longer than just a one word. In the case for example of a inverse of matrix it is impossible in reality to derive the formula in a quick way. I agree that mnemonics can play more important role in rather more complicated cases.... $\endgroup$
    – Widawensen
    Mar 28, 2017 at 8:52
  • $\begingroup$ Linear algebra does not have really any complicated formula that requires a device to remember. Actually a problem in mathematics is considered very simplified if one can reduce it to a linear algebra problem. Initially I found the formulas in Gram-Schmidt orthogonalization messy: but after realising that we repeatedly subtract projections to previously found partial set of orthonormal basis vectors it was easy. $\endgroup$ Mar 28, 2017 at 9:11
  • $\begingroup$ @PVanchinathan hmm, so my congratulations that you have such a level of understanding that you can easily grasp all details ... maybe practice makes a master... however sometimes we have not so much time as we should have.. $\endgroup$
    – Widawensen
    Mar 28, 2017 at 9:24
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    $\begingroup$ @Widawensen: Trying to learn mathematics by rote remembering of formulas is the slow way to go about it -- so slow that many who attempt it end up believing they are "bad at math" because they're stuck with an unfeasible approach. If you're pressed on time you should focus your efforts at understanding what you're doing, and do practice problems with a view to develop that understanding rather than just learning procedures by role. $\endgroup$ Mar 28, 2017 at 9:43

2 Answers 2


(Too long for a comment.)

I agree with some commenters here. Before you can build up muscle memory, it is often easier, or even faster, to derive what you need than to recall mnemonics. And derivation also makes you understand better. At least this is my own experience when linear algebra is concerned.

In recent years, the only formula that I almost need some mnemonics to help remembering is the formula for calculating the determinant of a block-$2\times2$ matrix when two adjacent sub-blocks commute. Consider $$ M=\pmatrix{A&B\\ C&D}, $$ where the four sub-blocks are square submatrices of identical sizes over some commutative ring. When some two adjacent sub-blocks of $M$ commute, we have (c.f. John Silvester, Determinants of Block Matrices) $$ \det M= \begin{cases} \det(AD-BC) & \text{ if } C,D \text{ commute},\\ \det(DA-CB) & \text{ if } A,B \text{ commute},\\ \det(DA-BC) & \text{ if } B,D \text{ commute},\\ \det(AD-CB) & \text{ if } A,C \text{ commute}. \end{cases} $$ This is analogous to the formula $\det\pmatrix{a&b\\ c&d}=ad-bc$, but care must be taken here because the orders of $A,B,C,D$ in the polynomials above (i.e. $AD-BC$ etc.) depend on which sub-block commutes with which.

Kind of messy, right? But if you truly understand how they are derived, you don't need any mnemonics. First, we use Gaussian elimination to eliminate the off-diagonal block among the pair of commuting sub-blocks. E.g. in the first case above, i.e. when $C$ and $D$ commute, we have $$ \pmatrix{A&B\\ C&D}\pmatrix{D&0\\ -C&I}=\pmatrix{AD-BC&B\\ 0&D}.\tag{1} $$ Take determinants on both sides, we get $\det(M)\det(D)=\det(AD-BC)\det(D)$. Cancelling out $\det(D)$, we get the result.

At this point, the derivation still looks tedious. However, note that in our derivation, the second block column of $(1)$ does not really matter to our end result. So, to find the right polynomial we need, all we only need to calculate $$ \pmatrix{A&B\\ C&D}\pmatrix{D\\ -C}. $$ In other words, when we have a row of commuting sub-blocks, we use a block column vector to kill off the off-diagonal commuting block ($C$ in this example), and the only thing that you need to memorise is the following:

It is the off-diagonal commuting sub-block that has a negative sign in the killer block vector.

With this in mind, it is now dead easy to see what polynomial to use in each of the above four cases: $$ \begin{cases} \pmatrix{A&B\\ C&D}\pmatrix{D\\ -C}=\pmatrix{AD-BC\\ 0} & \text{ if } C,D \text{ commute},\\ \\ \pmatrix{A&B\\ C&D}\pmatrix{-B\\ A}=\pmatrix{0\\ DA-CB} & \text{ if } A,B \text{ commute},\\ \\ \pmatrix{D&-B}\pmatrix{A&B\\ C&D}=\pmatrix{DA-BC&0} & \text{ if } B,D \text{ commute},\\ \\ \pmatrix{-C&A}\pmatrix{A&B\\ C&D}=\pmatrix{0&AD-CB} & \text{ if } A,C \text{ commute}. \end{cases} $$

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    $\begingroup$ You have found convincing example where understanding is better than some abstract mnemonic and in this case is even hard to find a mnemonic for the mentioned statement to memorize: .... offDiaComBloNS? $\endgroup$
    – Widawensen
    Mar 28, 2017 at 11:35
  • $\begingroup$ @Widawensen In a brief period of time, I did use the phrase "column first, row second" to memorise the ordering. If the two commuting blocks appear on a column, put them first in the multiplicands; if they lie on a row, put them after the non-commuting ones. That did help me to decide the ordering quickly, but I couldn't get rid of the worry that if I had remembered the formula wrongly. In contrast, thinking in terms of Gaussian elimination is much more reassuring. $\endgroup$
    – user1551
    Mar 28, 2017 at 11:49

I made a mnemotechnic (quick and dirty attached in the image) for remembering the dimensions of the iMage space & Null space and it's relation to matrix vector multiplication of an M x N matrix. I plan on making a more beautiful LaTeX version when I find the time.

Mnemotechnic for dimensions and null space image space (first post so link instead of image...)


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