Determine for what real values of x the matrix A is positive semidefinite.
Given the matrix $$A=begin{bmatrix} 1 & (x+1) & 1 \
(x+1) & 1 & (x+1)\
1 & (x+1) & 1
end{bmatrix}$$
Determine for what real values of x the matrix A is positive semidefinite.
So, I tried to find the eigenvalues of A and got the expression $$lambda (-lambda^2+3lambda+(2x^2+4x))=0 Rightarrow lambda=0 textrm{or} -lambda^2+3lambda+(2x^2+4x)=0$$
In the last equation I have $lambda=frac{3pmsqrt{8x^2+16x+9}}{2}$
Q: This is the best answer I can find as a function of $xinmathbb{R}$? How to write the range where $lambdageq0$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 13:49
Juliana de Souza
656
add a comment |
Given the matrix $$A=begin{bmatrix} 1 & (x+1) & 1 \
(x+1) & 1 & (x+1)\
1 & (x+1) & 1
end{bmatrix}$$
Determine for what real values of x the matrix A is positive semidefinite.
So, I tried to find the eigenvalues of A and got the expression $$lambda (-lambda^2+3lambda+(2x^2+4x))=0 Rightarrow lambda=0 textrm{or} -lambda^2+3lambda+(2x^2+4x)=0$$
In the last equation I have $lambda=frac{3pmsqrt{8x^2+16x+9}}{2}$
Q: This is the best answer I can find as a function of $xinmathbb{R}$? How to write the range where $lambdageq0$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 13:49
Juliana de Souza
656
add a comment |
Given the matrix $$A=begin{bmatrix} 1 & (x+1) & 1 \
(x+1) & 1 & (x+1)\
1 & (x+1) & 1
end{bmatrix}$$
Determine for what real values of x the matrix A is positive semidefinite.
So, I tried to find the eigenvalues of A and got the expression $$lambda (-lambda^2+3lambda+(2x^2+4x))=0 Rightarrow lambda=0 textrm{or} -lambda^2+3lambda+(2x^2+4x)=0$$
In the last equation I have $lambda=frac{3pmsqrt{8x^2+16x+9}}{2}$
Q: This is the best answer I can find as a function of $xinmathbb{R}$? How to write the range where $lambdageq0$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 13:49
Juliana de Souza
656
Given the matrix $$A=begin{bmatrix} 1 & (x+1) & 1 \
(x+1) & 1 & (x+1)\
1 & (x+1) & 1
end{bmatrix}$$
Determine for what real values of x the matrix A is positive semidefinite.
So, I tried to find the eigenvalues of A and got the expression $$lambda (-lambda^2+3lambda+(2x^2+4x))=0 Rightarrow lambda=0 textrm{or} -lambda^2+3lambda+(2x^2+4x)=0$$
In the last equation I have $lambda=frac{3pmsqrt{8x^2+16x+9}}{2}$
Q: This is the best answer I can find as a function of $xinmathbb{R}$? How to write the range where $lambdageq0$?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 13:49
Juliana de Souza
656
asked Dec 2 '18 at 13:49
Juliana de Souza
656
edited Dec 2 '18 at 13:57
asked Dec 2 '18 at 13:49
Juliana de Souza
656
asked Dec 2 '18 at 13:49
Juliana de Souza
656
asked Dec 2 '18 at 13:49
Juliana de Souza
656
656
add a comment |
add a comment |
1 Answer
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We have
- $det(1)=1$
- $detbegin{bmatrix} 1 & (x+1) \
(x+1) & 1 end{bmatrix}=1-(x+1)^2$ - $det A=0$
then by Sylvester criterion we need
$$1-(x+1)^2 > 0 iff x(x+2)<0 quad -2<x<0$$
and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.
answered Dec 2 '18 at 13:55
gimusi
1
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We have
- $det(1)=1$
- $detbegin{bmatrix} 1 & (x+1) \
(x+1) & 1 end{bmatrix}=1-(x+1)^2$ - $det A=0$
then by Sylvester criterion we need
$$1-(x+1)^2 > 0 iff x(x+2)<0 quad -2<x<0$$
and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.
answered Dec 2 '18 at 13:55
gimusi
1
add a comment |
We have
- $det(1)=1$
- $detbegin{bmatrix} 1 & (x+1) \
(x+1) & 1 end{bmatrix}=1-(x+1)^2$ - $det A=0$
then by Sylvester criterion we need
$$1-(x+1)^2 > 0 iff x(x+2)<0 quad -2<x<0$$
and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.
answered Dec 2 '18 at 13:55
gimusi
1
add a comment |
We have
- $det(1)=1$
- $detbegin{bmatrix} 1 & (x+1) \
(x+1) & 1 end{bmatrix}=1-(x+1)^2$ - $det A=0$
then by Sylvester criterion we need
$$1-(x+1)^2 > 0 iff x(x+2)<0 quad -2<x<0$$
and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.
answered Dec 2 '18 at 13:55
gimusi
1
We have
- $det(1)=1$
- $detbegin{bmatrix} 1 & (x+1) \
(x+1) & 1 end{bmatrix}=1-(x+1)^2$ - $det A=0$
then by Sylvester criterion we need
$$1-(x+1)^2 > 0 iff x(x+2)<0 quad -2<x<0$$
and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.
answered Dec 2 '18 at 13:55
gimusi
1
edited Dec 2 '18 at 14:24
answered Dec 2 '18 at 13:55
gimusi
1
answered Dec 2 '18 at 13:55
gimusi
1
answered Dec 2 '18 at 13:55
gimusi
1
1
add a comment |
add a comment |
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