Why can't we define an eigenvalue of a matrix as any scalar value?

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    Eigenvectors Matrix
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Homework Statement
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Relevant Equations
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For this,
1682737134968.png

Dose anybody please know why we cannot say ##\lambda = 1## and then ##1## would be the eigenvalue of the matrix?

Many thanks!
 
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The result of the multiplication is ##\begin{bmatrix} 1 \\ 5 \end{bmatrix}##, not ##\begin{bmatrix} \lambda \\ 0 \end{bmatrix}##, so it doesn't matter what the value of ##\lambda## is.
 
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ChiralSuperfields said:
Dose anybody please know why we cannot say λ=1 and then 1 would be the eigenvalue of the matrix?
"Dose" -- an amount of medicine.
"Does" -- third person singular conjugation of the infinitive verb "to do."

An eigenvalue ##\lambda## is a number such that for an eigenvector x, ##A\mathbf x = \lambda \mathbf x##.

For the matrix you asked about ##\begin{bmatrix}1 & 6 \\ 5 & 2\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \begin{bmatrix}1 \\5 \end{bmatrix} \ne \lambda \begin{bmatrix}1 \\ 0 \end{bmatrix}## for any value of ##\lambda##.
 
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Thank you for your replies @FactChecker and @Mark44!

Sorry I still don't I understand. I'll try to explain what my understanding is so that any misconception can be exposed. ##\lambda## is the constant in front that is factor out of the column vector ##\vec x## which is called the eigenvalue. For examples 1 and 2 below, the constant multiplied to the column vector is ##\lambda = 7 , -4## respectively.
1682746269493.png


However, for this example,

##\begin{bmatrix}1 & 6 \\ 5 & 2\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \begin{bmatrix}1 \\5 \end{bmatrix}##, why can't we factor out a 1 from the column vector to get ##\begin{bmatrix}1 & 6 \\ 5 & 2\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} = 1 \begin{bmatrix}1 \\5 \end{bmatrix}##.

According to the textbook, ##\lambda## can be any real number, so why can't ##1## be an eigenvalue?

Many thanks!
 
Mark44 said:
An eigenvalue ##\lambda## is a number such that for an eigenvector x, ##A\mathbf x = \lambda \mathbf x##.
You didn't read what I wrote in my previous post carefully enough. An eigenvalue is closely associated with a specific eigenvector. In the equation above, x is an eigenvector that appears on both sides of the equation. For an eigenvalue/eigenvector pair, multiplication of the vector by the matrix produces a value that is a scalar multiple (i.e., the eigenvalue) of that same vector.
ChiralSuperfields said:
However, for this example,
##\begin{bmatrix}1 & 6 \\ 5 & 2\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \begin{bmatrix}1 \\5 \end{bmatrix}##, why can't we factor out a 1 from the column vector to get ##\begin{bmatrix}1 & 6 \\ 5 & 2\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} = 1 \begin{bmatrix}1 \\5 \end{bmatrix}##.
Because ##\begin{bmatrix}1 \\ 0 \end{bmatrix}## isn't the vector that appears on both sides of the equation.
 
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Mark44 said:
You didn't read what I wrote in my previous post carefully enough. An eigenvalue is closely associated with a specific eigenvector. In the equation above, x is an eigenvector that appears on both sides of the equation. For an eigenvalue/eigenvector pair, multiplication of the vector by the matrix produces a value that is a scalar multiple (i.e., the eigenvalue) of that same vector.

Because ##\begin{bmatrix}1 \\ 0 \end{bmatrix}## isn't the vector that appears on both sides of the equation.
Oh, thank you @Mark44! I see now. Sorry I forgot that the column vector has to be on both sides.
 
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