0
$\begingroup$

I was dealing with projection matrix topic.

It states the matrix of projection on the line through $a=(a_1,a_2,a_3)$ is given by

$P=\frac{aa^T}{a^Ta}$

Here while solving the sum they used $a$ as column vector and hence $a^T$ is a row vector. So numerator they were getting a matrix.

But my question is, can I use $a$ as row vector then $a^T$ will be the column vector but then I will not get matrix in the numerator but in the denominator, will it be still right or make any sense. Or we are always suppose to take $a$ as column vector here.

Thanks in advance...

$\endgroup$
3
  • $\begingroup$ Always use a column vector for this equation. You won't get the right answer otherwise. $\endgroup$ Commented Jul 14, 2024 at 17:12
  • $\begingroup$ But this cannot be the reason to choose column vector just to obtain right answers. Is it related to the derivation of the formula where column vector must be considered for a? $\endgroup$ Commented Jul 14, 2024 at 17:48
  • $\begingroup$ The formula $\frac{a^Ta}{aa^T}$ works for row vectors. However column vectors are the standard in the literature. You can do everything with row vectors instead but we choose not to do that mostly so we can write $Ax$ instead of $xA$ for matrix acting on $x$. $\endgroup$ Commented Jul 14, 2024 at 22:02

2 Answers 2

Reset to default
1
$\begingroup$

The simple answer, no it doesn't make sense in this case. Because what does it mean to have a matrix in the denominator? You can interpret it benevolently as the inverse of the matrix, but this does not always work, as not every matrix is reversible.

$\endgroup$
1
  • $\begingroup$ In this case it essentially never works because $a^Ta$ will be rank $1$. $\endgroup$ Commented Jul 14, 2024 at 17:23
0
$\begingroup$

In short, the projection matrix of $A$ still makes sense no matter $A$ is a column vector, a row vector, or a matrix. But $P$ may or may not be written in the fraction form.

In general, the projection matrix for projecting onto the columns space of real matrix $A$ is $P=A(A^TA)^{-}A^T$, where $(A^TA)^-$ is any generalized inverse of $A^TA$.

  • If $A$ is a non-zero column vector $a$, this is identical to the $P=\frac{aa^T}{a^Ta}$ in the question. For any compatible column vector $b$, $Pb$ will be proportional to $a$, i.e., in the column space of $a$.
  • If $A$ is a zero column vector $a$, then $P$ is just a zero matrix, but the fraction form in the question does not apply, since we cannot divide by zero. Now $Pb$ is just $a=\mathbf{0}$, i.e., a zero vector, because the column space of $A$ is just the single-element set of a zero vector.
  • If $A$ is a non-zero row vector, what we get is just $P=A(A^TA)^{-}A^T=1$. Here a generalized inverse has to be used (except when $A$ is $1\times1$). The fraction form in the question no longer applies, since a matrix should not show up in the denominator (except when it is $1\times1$). For any other real value $b$, its projection is $Pb=b$ itself, not changing anything, because the column space of $A$ is just $\mathbb{R}$, in which $b$ lies.
  • If $A$ is a zero row vector, now we have $P=0$, and $Pb=0$. It still makes sense because the column space of a row of zeros is the set $\lbrace 0 \rbrace$.
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.