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Determinant is linear as a function of each of the rows of the matrix.

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Determinant is linear as a function of each of the rows of the matrix.

2

$begingroup$

Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.

I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$

Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?

linear-algebra matrices determinant

share|cite|improve this question

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
  $endgroup$
  – StammeringMathematician
  57 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
  $endgroup$
  – copper.hat
  57 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.

I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$

Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?

linear-algebra matrices determinant

share|cite|improve this question

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
  $endgroup$
  – StammeringMathematician
  57 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
  $endgroup$
  – copper.hat
  57 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.

I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$

Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?

linear-algebra matrices determinant

share|cite|improve this question

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

$endgroup$

share|cite|improve this question

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

share|cite|improve this question

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

edited 1 hour ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

2 Answers
2

active

oldest

votes

3

$begingroup$

If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

add a comment | 

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3

$begingroup$

If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

add a comment | 

3

$begingroup$

If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

add a comment | 

$begingroup$

If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

add a comment | 

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

share|cite|improve this answer

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639

answered 59 mins ago

TomGrubbTomGrubb

11.2k11639