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Suppose that $P$ is a $4 \times 5$ matrix such that every solution of the equation $\text{Px=0}$ is a scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$. The rank of $P$ is __________

These question can be solved using rank-nullity theorem, rank+nullity=no of variables.here they have mentioned all solution of the equation Px=0 are the scalar multiple of [2 5 4 3 1]^T.so,we can understand that n-1 solutions are dependent while one and only one solution is independent.And by the definition, nullity=number of independent solution,so nullity=1.no of variables is nothing but the no of columns in owr 4×5 matrix which is 5. Therefore rank=5-1=4.hope these is helpful.

5

1^{st} thing: Rank + Nulity = No. of unknowns (no. of columns )

Nulity refers Nullspace of the real m X n matrix

Nullspace contains those vectors which satisfies the Homogeneous equation

i.e If A is the Matrix then Nullspace of A will have vector(s) “x” such that Ax=0

Evetually Nullspace of a Matrix = Span(all vectors “x”)

Coming to the Question,

Here the x (vector(s)) are all Lr.dependent ,

So, Null space = 1

=nulity =1

Rank = 5-1

check this: https://brilliant.org/wiki/rank-nullity-theorem/

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9 votes

Best answer

Every solution to $Px = 0$ is scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$, It means out of 5 column vectors of matrix $P$, $4$ are linearly independent as we have only one line in NULL Space (along the given vector).

Rank is nothing but the number of linearly independent column vectors in a matrix which is $4$ here.

Rank is nothing but the number of linearly independent column vectors in a matrix which is $4$ here.

edited
Aug 26, 2021
by Divyanshu Shukla

If we consider $P=\left \{ {C_1},{C_2},{C_3},{C_4},{C_5}\right \}$ and $X=\begin{bmatrix} x_1 &x_2 & x_3 & x_4 & x_5 \end{bmatrix}^T$

Since $P X = 0$

Then, $x_1C_1+x_2C_2+x_3C_3+x_4C_4+x_5C_5=0$

$\left ( Given, x_1=2k, x_2=5k, x_3=4k,x_4=3k,x_5=k \right )$

So, $2C_1+5C_2+4C_3+3C_4+C_5=0$

$C5=-2C_1-5C_2-4C_3-3C_4$

This proves that one column vector is dependent on the other $4$ independent column vectors.

Rank = number of linearly independent column vectors in a matrix which is $4$ here.

Other methods to solve this problem:

The dimension of the null space of matrix P is called the nullity of P.

Let say, $u=\begin{bmatrix} 2 &5 &4 & 3 & 1 \end{bmatrix}^T$

Since every other solution can be expressed in the form of given vector, so we can write $X=k\cdot u$. This means u form the basis of null$(P)$ and null space has dimension 1. Hence nullity$(P)=1$,

Or,

nullity is the no of free variables in the solution of $PX=0$.

Here, $X=\begin{bmatrix} 2k \\ 5k \\ 4k \\ 3k \\k \end{bmatrix}$

Clearly, $x_1,x_2,x_3,x_4$ are dependent on $x_5$. Only 1 free variable in solution.

so nullity$(P)$ =1.

By rank-nullity theorem,

rank$(P)+$ nullity$(P)=$ no of column of P

$r+1=5 \rightarrow r=4$

(Correct me if something wrong).

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0

7 votes

Rank $+$ Nullity = Number of Columns

Here, Nullity is $1$. (Nullity is the dimension of the null space)

Rank : $5\ – 1 = 4$

Here, Nullity is $1$. (Nullity is the dimension of the null space)

Rank : $5\ – 1 = 4$

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@Anuj,

Nullity is the dimension of the null space

in other words,

Nullity of a matrix

Ais no. of possible column vectorsXsatisfying the equationAX = 0

In the given question, we are interested in finding the nullity of P. So we need to find out the number of possible column vectors **X** satisfying the equation **PX = 0.** In the question, it is also given that

every solution of the equation PX=0 is a scalar multiple of [25431]

$^T$

*this means that X is of the form k* [25431]*$^T$ where k is a scalar. Clearly, the Null Space of P contains only 1 vector. Hence Nullity is 1.*

*You can also refer to this video for better understanding.*

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