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[EDIT: added the case of negative eigenvalues with larger (odd) multiplicity.]

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a real negative eigenvalue with odd multiplicity, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, by using a Jordan form one sees that if $A$ has a Jordan form with eigenvalues $\lambda_1,\dots,\lambda_s$, then $B = \exp(A)$ has a Jordan form with eigenvalues $e^{\lambda_1},\dots,e^{\lambda_s}$, the same block sizes and eigenvectors. Take now an eigenpair $Bv = v\lambda$ with $\lambda>0$, and take $v$ to be real (which is always possible). Then $A$ mus have an eigenvalue $\log \lambda$, with eigenvector $v$, and that is impossible since $Av$ is real and $v\log \lambda$ is not.

To conclude, we must note that having a negative real eigenvalue with odd multiplicity is an open property: if $B$ has a such an eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has the same property. Indeed, the characteristic polynomial $p_B(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$. So we can find $a<b<0$ such that $p_B(a)$ and $p_B(b)$ have opposite signs. A sufficiently close matrix $C$ will have a sufficiently close characteristic polynomial, hence $p_C(a)$ and $p_C(B)$ have opposite signs, and $C$ has an eigenvalue in $[a,b]$ with odd multiplicity, too.

[EDIT: added and then removed a stronger argument that did not work.]

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a simple negative eigenvalue, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a simple real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, note that we can find a real eigenvector $v$ for $B$. Any matrix logarithm must have an eigenvalue $\log \lambda \not\in\mathbb{R}$ with an associated real eigenvector $v$, as can be seen with a Jordan form. That is impossible, since $Av$ must be real.

To conclude, we must note that having a negative real eigenvalue with odd multiplicity is an open property: if $B$ has a such an eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has the same property. Indeed, the characteristic polynomial $p_B(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$. So we can find $a<b<0$ such that $p_B(a)$ and $p_B(b)$ have opposite signs. A sufficiently close matrix $C$ will have a sufficiently close characteristic polynomial, hence $p_C(a)$ and $p_C(B)$ have opposite signs, and $C$ has an eigenvalue in $[a,b]$ with odd multiplicity, too.

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a real negative simple eigenvalue, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a simple real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, note that we can find a real eigenvector $v$ for $B$. Any matrix logarithm $A\in\mathbb{R}$ must have an eigenvalue $\log \lambda \not\in\mathbb{R}$ with the same real eigenvector $v$, as can be seen with a Jordan form. That is impossible, since $Av$ must be real.

To conclude, we must note that having a real simple eigenvalue is an open property: if $B$ has a real negative eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has a negative eigenvalue. This can be seen, for instance, by noting that the characteristic polynomial $p(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$.

[EDIT: added the case of negative eigenvalues with larger (odd) multiplicity.]

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a real negative eigenvalue with odd multiplicity, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, by using a Jordan form one sees that if $A$ has a Jordan form with eigenvalues $\lambda_1,\dots,\lambda_s$, then $B = \exp(A)$ has a Jordan form with eigenvalues $e^{\lambda_1},\dots,e^{\lambda_s}$, the same block sizes and eigenvectors. Take now an eigenpair $Bv = v\lambda$ with $\lambda>0$, and take $v$ to be real (which is always possible). Then $A$ mus have an eigenvalue $\log \lambda$, with eigenvector $v$, and that is impossible since $Av$ is real and $v\log \lambda$ is not.

To conclude, we must note that having a negative real eigenvalue with odd multiplicity is an open property: if $B$ has a such an eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has the same property. Indeed, the characteristic polynomial $p_B(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$. So we can find $a<b<0$ such that $p_B(a)$ and $p_B(b)$ have opposite signs. A sufficiently close matrix $C$ will have a sufficiently close characteristic polynomial, hence $p_C(a)$ and $p_C(B)$ have opposite signs, and $C$ has an eigenvalue in $[a,b]$ with odd multiplicity, too.

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a real negative simple eigenvalue, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can conclude.

A converse holds: suppose $B$ has a simple real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, note that we can find a real eigenvector $v$ for $B$. Any matrix logarithm $A\in\mathbb{R}$ must have an eigenvalue $\log \lambda \not\in\mathbb{R}$ with the same real eigenvector $v$, as can be seen with a Jordan form. That is impossible, since $Av$ must be real.

To conclude, we must note that having a real simple eigenvalue is an open property: if $B$ has a real negative eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has a negative eigenvalue. This can be seen, for instance, by noting that the characteristic polynomial $p(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$.

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a real negative simple eigenvalue, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a simple real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, note that we can find a real eigenvector $v$ for $B$. Any matrix logarithm $A\in\mathbb{R}$ must have an eigenvalue $\log \lambda \not\in\mathbb{R}$ with the same real eigenvector $v$, as can be seen with a Jordan form. That is impossible, since $Av$ must be real.

To conclude, we must note that having a real simple eigenvalue is an open property: if $B$ has a real negative eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has a negative eigenvalue. This can be seen, for instance, by noting that the characteristic polynomial $p(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$.