# Interesting/Useful tricks in linear algebra. [closed]

This may be an opinion based-based question, since everyone understands things differently, but what are some interesting/useful tricks in linear algebra?

For example, if we know that the determinant of a matrix is equal to 0 0 , then we that the matrix is not invertible, the rows and columns of the matrix are linearly dependent, and so on.

Are there any others, that would be useful to have on a cheatsheet for a linear algebra exam?

• That's actually not a trick... That's consequences (?) I guess or maybe property – UmbQbify Jul 1 '20 at 11:46
• There is a nice classification when a linear equation system has no solution, a unique solution or infinite many solutions. All you need is the ranks of the matrices $A$ and $A$ concanated with $b$ if $Ax=b$ is the system. – Peter Jul 1 '20 at 11:46
• @SkiMask How we call it does not matter , if it is useful. I see no problem with calling such things a "trick". "Tricks" almost always involve proven results. – Peter Jul 1 '20 at 11:48
• Personally, I like the cyclic property of the trace and the matrix determinant lemma. Here is an example. – Rodrigo de Azevedo Jul 1 '20 at 12:54
• This question is not only opinion-based, but way too broad in scope: If not narrowed down, “tricks” can be all sorts of things – from grand theorems, over well-known elementary facts to small notational devices. And I don’t think we need a list of well-known theorems and facts about linear algebra. So the only way this list could turn out useful in any way is if we were to limit ourselves to “tricks” of which we think that they are only little known. I also think, examples of applications should be provided. – k.stm Jul 1 '20 at 15:53

My top 3 (which many of my students keep forgetting/not understanding):

1. You can perform elementary row operations by multiplying a given matrix with the appropriate regular matrix.
2. Linear transformations and matrices are "the same thing" and questions can be translated between the two relatively easily.
3. You can define a linear transformation by defining it on a basis, no need to come up with a general formula.
• Hmm ... I recently also understood, thanks to 3blue1brown, that a matrix is essentially a linear transformation. However, the first point is quite interesting. – Ski Mask Jul 1 '20 at 12:10
• Could you elaborate the first point? – UmbQbify Jul 1 '20 at 12:59
• About #1: en.wikipedia.org/wiki/…. – YJT Jul 1 '20 at 13:03
1. Rank factorization, you can write any matrix $$\textbf{A}_{m\times n}=\textbf{B}\textbf{C}$$, where $$\rho(A)=r$$ and $$\textbf{B}$$ is an $$m\times r$$ and $$\textbf{C}$$ is an $$r\times n$$ matrix. If you use properly it is a very useful property of matrix. For positive definite matrix you can have $$\textbf{A}_{n\times n}=\textbf{B}^{T}\textbf{B}$$.

2. Block matrix, $$\textbf{A}= \left[\begin{matrix} \textbf{A}_{11} & \textbf{A}_{12} \\ \textbf{A}_{21} & \textbf{A}_{22} \\ \end{matrix}\right]$$, Now suppose we required $$\textbf{A}_{11}$$. then consider $$\textbf{B}=\left[\begin{matrix}\textbf{B}_1 \\ \textbf{B}_2\end{matrix}\right]$$ and $$\textbf{C}=\left[\begin{matrix}\textbf{C}_1 & \textbf{C}_2\end{matrix}\right]$$, then $$\textbf{A}=\left[\begin{matrix}\textbf{B}_1 \\ \textbf{B}_2\end{matrix}\right] \left[\begin{matrix}\textbf{C}_1 & \textbf{C}_2\end{matrix}\right] = \left[\begin{matrix} \textbf{B}_{1}\textbf{C}_1 & \textbf{B}_{1}\textbf{C}_2 \\ \textbf{B}_{2}\textbf{C}_1 & \textbf{B}_{2}\textbf{C}_2 \\ \end{matrix}\right]$$, therefore, $$\textbf{A}_{11} = \textbf{B}_{1}\textbf{C}_1$$.

3. Spectral decomposition of matrix $$\textbf{A}$$. That is $$\textbf{A}_{m\times n}=\textbf{B}^{-1}\textbf{D}\textbf{B}$$, where $$\textbf{D}=$$ diagonal matrix using eigen-value of $$\textbf{A}$$.

4. Some properties of kronecker product, such that $$(\textbf{A}\otimes \textbf{B})(\textbf{C}\otimes \textbf{D})=\textbf{AC}\otimes \textbf{BD}$$. Using this identity with spectral decompostion, you can observe many useful property, such as kronecker product of to positive definite matrix is positive definite.

5. Hadamard product. there are some claver use of Hadamard product to show a matrix positive definite. As a example, suppose $$\textbf{A}= \left[\begin{matrix} a & b \\ c & d \\ \end{matrix}\right]$$ and $$\textbf{B}= \left[\begin{matrix} e & f \\ g & h \\ \end{matrix}\right]$$, and $$\textbf{A}$$ and $$\textbf{B}$$ are positive definite matix. Then what can we say about the matrix $$\left[\begin{matrix} ae & bf \\ cg & dh \\ \end{matrix}\right]$$? Obviously it is a positive definite matrix, but how can we show that? Consider $$\textbf{A}\otimes \textbf{B}$$, this is positive definite matrix. now $$\textbf{Z}_{2\times 4}$$ a elimentary matrix. then $$\textbf{Z}(\textbf{A}\otimes \textbf{B})\textbf{Z}^{T}$$ is positive definite. you can choose $$\textbf{Z}$$ such a way that $$\textbf{Z}(\textbf{A}\otimes \textbf{B})\textbf{Z}^{T}=\left[\begin{matrix} ae & bf \\ cg & dh \\ \end{matrix}\right]$$ is positive definite.

Here, $$\left[\begin{matrix} ae & bf \\ cg & dh \\ \end{matrix}\right] = \textbf{A}\circ\textbf{B}$$, Hadamard product of $$\text{A}$$ and $$\textbf{B}$$.

last but not the least, There are some useful identity in linear algebra such as Woodbury identity ,Schur compliment etc. you can find useful.

• That's quite a nice list! – Ski Mask Jul 1 '20 at 15:31