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Let $A$ be a $n \times n$ invertible matrix, prove $\det(kA)=k^n\det A$. I really don't know where to start. Can someone give me a hint for this proof?

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What definition of det are you using? –  Chris Eagle Jul 12 '13 at 18:09
    
@BabakS.Sorry, it's a typo. Edited. –  user2185071 Jul 12 '13 at 18:12
    
What row operations will get you from $A$ to $kA$? –  Michael Jul 12 '13 at 18:12
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4 Answers

up vote 6 down vote accepted

First, let's recall what multiplication of a matrix by a scalar means: When we multiply $\;kA$, where $A$ is an $n\times n$ matrix, and $k$ a scalar, then every entry $a_{ij}$ of matrix $A$ is multiplied by $k$: i.e. $a_{ij}\mapsto ka_{ij}$ for each $a_{ij}$. That means for each row $i$ $0\leq i\leq n$, we can factor out $k$.

Now...Recall the elementary row operations you've learned, and how each one of them affects the determinant of the matrix on which it is operating. Specifically, when any one row is multiplied by the scalar $k$, the determinant of $A$ becomes $\;k\det A$. So given that scalar multiplication of a square $n\times n$ matrix is equivalent to "row operating" on $n$ rows (by multiplying each row by the scalar $k$), we can conclude that $$\large\det (kA) = \underbrace{k\cdot k\cdot\cdots \cdot k}_{\large n \;\text{times}}\det A = k^n\det A$$

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Clear as always +1 –  Amzoti Jul 13 '13 at 0:25
    
Clear and beautiful! –  Sami Ben Romdhane Mar 12 at 16:59
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Determinant is a multi-linear function, it means that $\det(x_1,\ldots,kx_i + y_i,\ldots,x_n) = k \det(x_1,\ldots, x_n) + \det(x_1,\ldots, y_i, \ldots,x_n)$, in which the i-th component is the i-th row of the matrix. by using $y_i = 0$ your desired result follows.

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Suppose that $v_1,\dots,v_n$ are the column vectors of the matrix $A$, then we can think of $\det$ instead of a function that maps one matrix to the reals as a function that maps $n$ vectors to the reals. Indeed, this is the way we define $\det$, it is a multilinear function of $n$ vectors! Now, being multilinear means that for each $i$ with $1\leq i\leq n$ we have:

$$\det(v_1\dots,kv_i,\dots,v_n)=k\det(v_1,\dots,v_i,\dots,v_n)$$

$$\det(v_1,\dots,u+w,\dots,v_n)=\det(v_1,\dots,u,\dots,v_n)+\det(v_1,\dots,w,\dots,v_n)$$

We simply need to use the first property. If $A=\begin{pmatrix}v_1 & \cdots & v_n\end{pmatrix}$ is your matrix, then by the usual definition of scalar multiplication for matrices we have $kA=\begin{pmatrix}kv_1 & \cdots & kv_n\end{pmatrix}$. In that case, we have that:

$$\det(kA)=\det(kv_1,\dots,kv_n)=k^n\det(v_1,\dots,v_n)=k^n\det(A)$$

Where $k^n$ appears because by multilinearity, for each index, one $k$ will be pulled out the $\det$. Pulling $n$ times the $k$ out, we get $k^n$ as desired.

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Just another way of doing it:

If $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of a matrix, then $\det(A)=\lambda_1\times\lambda_2\times\cdots\times\lambda_n$.

Let $\lambda$ be any eigenvalue of $A$ and $x$ be its corresponding eigenvector. Now, $Ax=\lambda x$, then it is easy to see that $k\lambda$ is the eigenvalue of $kA$ with the same eigenvector. Now, using the fact that each eigenvalue is being multiplied by $k$ and the determinant is product of eigenvalues, your proof follows.

I do wish to point out that the only reason I put this proof in is for fun. It is just another way of doing it. What you should be doing ideally is to work from the definition of a determinant taught to you.

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